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Himpunan penyelesaian persamaan cos2x+3cosx+2=0untuk0∘≤x≤360∘...

Pertanyaan

Himpunan penyelesaian persamaan cos invisible function application 2 x plus 3 cos invisible function application x plus 2 equals 0 space u n t u k space 0 degree less or equal than x less or equal than 360 degree...

  1. left curly bracket 60 degree comma space 120 degree comma space 270 degree right curly bracket

  2. left curly bracket 120 degree comma space 240 degree comma space 270 degree right curly bracket

  3. left curly bracket 90 degree comma space 240 degree comma space 270 degree right curly bracket

  4. left curly bracket 120 degree comma space 180 degree comma space 240 degree right curly bracket

  5. left curly bracket 120 degree comma space 150 degree comma space 270 degree right curly bracket

Pembahasan Soal:

I n g a t space 2 space m a c a m space i d e n t i t a s space cos invisible function application 2 x  A right parenthesis cos invisible function application 2 x equals 2 cos squared invisible function application x minus 1  B right parenthesis cos invisible function application 2 x equals 1 minus 2 sin squared invisible function application x    P a k a i space i d e n t i t a s space p e r t a m a space u n t u k space m e n y e l e s a i k a n space p e r s a m a a n space d i a t a s space  c o s invisible function application 2 x plus 3 c o s invisible function application x plus 2 equals 0  left parenthesis 2 c o s squared invisible function application x minus 1 right parenthesis plus 3 c o s invisible function application x plus 2 equals 0  2 c o s squared invisible function application x plus 3 c o s invisible function application x plus 1 equals 0  left parenthesis 2 c o s invisible function application x plus 1 right parenthesis left parenthesis c o s invisible function application x plus 1 right parenthesis equals 0  c o s invisible function application x equals negative 1 half space a t a u space c o s invisible function application x equals negative 1    C o sin u s space b e r n i l a i space n e g a t i v e space u n t u k space k u a d r a n space 2 space left parenthesis 180 minus x right parenthesis space d a n space 3 space left parenthesis 180 plus x right parenthesis  A right parenthesis space cos invisible function application x equals negative 1 half rightwards double arrow 180 minus 60 equals 120 space d a n space 180 plus 60 equals 240  B right parenthesis space cos invisible function application x equals negative 1 rightwards double arrow x equals 180  J a d i space h i m p u n a n space p e n y e l e s a i a n n y a space a d a l a h space left curly bracket 120 to the power of 0 comma 180 to the power of 0 comma 240 to the power of 0 right curly bracket space left parenthesis D right parenthesis

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 04 Oktober 2021

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Pertanyaan yang serupa

Himpunan penyelesaian persamaan trigonometri cos2x∘−sinx∘=0untuk0≤x≤180 adalah....

Pembahasan Soal:

cos invisible function application 2 x minus sin invisible function application x equals 0  1 minus 2 sin squared invisible function application x minus sin invisible function application x equals 0  2 sin squared invisible function application x plus sin invisible function application x minus 1 equals 0  open parentheses 2 sin invisible function application x minus 1 close parentheses open parentheses sin invisible function application x plus 1 close parentheses equals 0  D i d a p a t space sin invisible function application x equals 1 half space a t a u space sin invisible function application x equals negative 1  sin invisible function application x equals 1 half rightwards double arrow x equals 30 degree comma space 150 degree  s i n invisible function application x equals negative 1 rightwards double arrow x equals 270 degree space left parenthesis t i d a k space m e m e n u h i space r a n g e space 0 less or equal than x less or equal than 180 right parenthesis  M a k a space h i m p u n a n space p e n y e l e s a i a n space y a n g space m e m e n u h i space a d a l a h space left curly bracket 30 degree comma 150 degree right curly bracket

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Roboguru

Himpunan penyelesaian cos 2x = 3 cos x - 2 untuk 0≤x≤2π adalah ....

Pembahasan Soal:

cos2x=3cosx - 2

cos2x - 3cosx + 2=0

left parenthesis 2 cos squared x minus 1 right parenthesis minus 3 cos x plus 2 equals 0  2 cos squared x minus 3 cos x plus 1 equals 0  left parenthesis 2 cos x minus 1 right parenthesis left parenthesis cos x minus 1 right parenthesis equals 0  cos x equals 1 half space a t a u space cos x equals 1  P e n y e l e s a i a n space I colon  cos x equals 1 half  cos x equals cos straight pi over 3  x equals straight pi over 3  cos x equals 1 half  cos x equals cos fraction numerator 5 straight pi over denominator 3 end fraction  x equals fraction numerator 5 straight pi over denominator 3 end fraction  P e n y e l e s a i a n space I I colon  cos x equals 1  cos x equals cos 0  x equals 0  d a n  cos x equals 1  cos x equals cos 2 straight pi  straight x equals 2 straight pi  Jadi space himpunan space penyelesaian space adalah

left curly bracket 0 comma straight pi over 3 comma space 5 over 3 straight pi comma space 2 straight pi right curly bracket

 

0

Roboguru

Himpunan penyelesaian dari persamaan cos 2x- 3 cosx+ 2 = 0, 0∘≤x≤360∘ adalah...

Pembahasan Soal:

Ingat!
Persamaan trigonometri dengan bentuk:

  • cos x = cos begin mathsize 12px style alpha end style maka begin mathsize 12px style x equals plus-or-minus alpha plus k cross times 360 degree end style, untuk k = 0, 1, 2, 3, ...
  • begin mathsize 12px style cos space 2 x equals 2 space cos squared space x minus 1 end style

cos 2x- 3 cosx+ 2 = 0, begin mathsize 12px style 0 degree less or equal than x less or equal than 360 degree end style, maka:

begin mathsize 12px style 2 space cos squared space x minus 1 minus 3 space cos space x plus 2 equals 0  2 space cos squared space x minus 3 space cos space x plus 1 equals 0  open parentheses 2 space cos space x minus 1 close parentheses open parentheses cos space x minus 1 close parentheses equals 0 end style

Dari persamaan di atas, diperoleh:

  • 2 cos x - 1 = 0 atau cos x - 1 = 0
  • cos x = begin mathsize 12px style 1 half end style atau cos x = 1
  • cos x = cos begin mathsize 12px style 60 degree end style atau cos x = cos begin mathsize 12px style 0 degree end style

Penyelesaian dari persamaan trigonometri tersebut:

cos x = cos begin mathsize 12px style 60 degree end style

x = begin mathsize 12px style 60 degree plus k cross times 3 60 degree end style (untuk k = 0)

x = begin mathsize 12px style 60 degree end style

x = begin mathsize 12px style negative 60 degree plus k cross times 3 60 degree end style (untuk k = 1)

x = begin mathsize 12px style 300 degree end style

 

cos x = cos begin mathsize 12px style 0 degree end style

x = begin mathsize 12px style 0 degree plus k cross times 360 degree end style (untuk k = 0 dan 1)

x = begin mathsize 12px style 0 degree end style dan begin mathsize 12px style 360 degree end style

 

Jadi, himpunan penyelesaian dari persamaan cos 2x- 3 cos x + 2 = 0; untuk begin mathsize 12px style 0 degree less or equal than x less or equal than 360 degree end style adalah begin mathsize 12px style open curly brackets 0 degree comma space 60 degree comma space 300 degree comma space 360 degree close curly brackets end style

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Himpunan penyelesaian persamaan cos 2x - 3 cos x + 2 = 0 untuk 0 ≤ x  2π adalah...

Pembahasan Soal:

Ingat! 

begin mathsize 12px style cos space 2 x equals 2 space cos squared space x minus 1  cos space 2 x space minus space 3 space cos space x space plus space 2 space equals space 0  open parentheses 2 space cos squared space x minus 1 close parentheses minus 3 space cos space x plus 2 equals 0  open parentheses 2 space cos squared space x close parentheses minus space 3 space cos space x plus 1 equals 0 end style

cos x = begin mathsize 12px style 1 half end style atau cos x = 1

Diperoleh:

cos x = begin mathsize 12px style 1 half end style begin mathsize 12px style left right double arrow end style x = begin mathsize 12px style 60 degree comma space 300 degree end style

cos x = 1 begin mathsize 12px style left right double arrow end style x = begin mathsize 12px style 0 degree comma space 360 degree end style

Jadi, himpunan penyelesaiannya adalah begin mathsize 12px style open curly brackets 0 comma space straight pi over 3 comma space 5 over 3 straight pi comma space 2 straight pi close curly brackets end style.

0

Roboguru

Himpunan nilai x yang memenuhi 2sin(2x+15∘)+2cos(2x+15∘)=sin(2x+15∘)−cos(2x+15∘)1​  dengan −90∘≤x≤90∘ adalah ....

Pembahasan Soal:

Perhatikan bahwa

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 sin space left parenthesis 2 x plus 15 degree right parenthesis plus 2 cos space left parenthesis 2 x plus 15 degree right parenthesis end cell equals cell fraction numerator 1 over denominator sin space left parenthesis 2 x plus 15 degree right parenthesis minus cos space left parenthesis 2 x plus 15 degree right parenthesis end fraction end cell row cell 2 open parentheses sin space left parenthesis 2 x plus 15 degree right parenthesis plus cos space left parenthesis 2 x plus 15 degree right parenthesis close parentheses end cell equals cell fraction numerator 1 over denominator sin space left parenthesis 2 x plus 15 degree right parenthesis minus cos space left parenthesis 2 x plus 15 degree right parenthesis end fraction end cell end table

Untuk menghilangkan penyebut pada persamaan tersebut bisa dikalikan kedua ruasnya dengan begin mathsize 14px style sin space left parenthesis 2 x plus 15 degree right parenthesis minus cos space left parenthesis 2 x plus 15 degree right parenthesis end style sehingga menjadi

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 open parentheses sin space open parentheses 2 x plus 15 degree close parentheses plus cos space open parentheses 2 x plus 15 degree close parentheses close parentheses open parentheses sin space open parentheses 2 x plus 15 degree close parentheses minus cos space open parentheses 2 x plus 15 degree close parentheses close parentheses end cell equals 1 row cell sin squared space open parentheses 2 x plus 15 degree close parentheses minus cos squared space open parentheses 2 x plus 15 degree close parentheses end cell equals cell 1 half end cell row cell negative open parentheses cos squared space open parentheses 2 x plus 15 degree close parentheses minus sin squared space open parentheses 2 x plus 15 degree close parentheses close parentheses end cell equals cell 1 half end cell row cell cos squared space open parentheses 2 x plus 15 degree close parentheses minus sin squared space open parentheses 2 x plus 15 degree close parentheses end cell equals cell negative 1 half end cell end table

Ingat bahwa cos space 2 x equals cos squared space x minus sin squared space x sehingga cos squared space open parentheses 2 x plus 15 degree close parentheses minus sin squared space open parentheses 2 x plus 15 degree close parentheses equals cos space 2 open parentheses 2 x plus 15 degree close parentheses equals cos space open parentheses 4 x plus 30 degree close parentheses. Selanjutnya, akan diperoleh

cos space open parentheses 4 x plus 30 degree close parentheses equals negative 1 half

Diketahui bahwa cos space 120 degree equals negative 1 half, maka diperoleh

cos space open parentheses 4 x plus 30 degree close parentheses equals cos space 120 degree

Persamaan cos space x equals cos space a terpenuhi oleh

x equals a plus k times 360 degree

atau

x equals negative a plus k times 360 degree

dengan k element of straight natural numbers.

Oleh karena itu, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x plus 30 degree end cell equals cell 120 degree plus k times 360 degree end cell row cell 4 x end cell equals cell 90 degree plus k times 360 degree end cell row x equals cell 22 comma 5 degree plus k times 90 degree end cell end table

Jika k equals negative 2, maka x equals 22 comma 5 degree plus open parentheses negative 2 close parentheses times 90 degree equals 22 comma 5 degree minus 180 degree equals negative 157 comma 5 degree. Nilai tersebut tidak memenuhi syarat x sehingga untuk nilai k makin kecil pasti tidak memenuhi syarat x juga.

Jika k equals negative 1, maka x equals 22 comma 5 degree plus open parentheses negative 1 close parentheses times 90 degree equals 22 comma 5 degree minus 90 degree equals negative 67 comma 5 degree. Nilai tersebut memenuhi syarat x.

Jika k equals 0, maka x equals 22 comma 5 degree plus open parentheses 0 close parentheses times 90 degree equals 22 comma 5 degree plus 0 degree equals 22 comma 5 degree. Nilai tersebut memenuhi syarat x.

Jika k equals 1, maka x equals 22 comma 5 degree plus open parentheses 1 close parentheses times 90 degree equals 22 comma 5 degree plus 90 degree equals 112 comma 5 degreeNilai tersebut tidak memenuhi syarat x sehingga untuk nilai k makin besar pasti tidak memenuhi syarat x juga.

Selanjutnya, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x plus 30 degree end cell equals cell negative 120 degree plus k times 360 degree end cell row cell 4 x end cell equals cell negative 150 degree plus k times 360 degree end cell row x equals cell negative 37 comma 5 degree plus k times 90 degree end cell end table

Jika k equals negative 1, maka x equals negative 37 comma 5 degree plus open parentheses negative 1 close parentheses times 90 degree equals negative 37 comma 5 degree minus 90 degree equals negative 127 comma 5 degree. Nilai tersebut tidak memenuhi syarat x sehingga untuk nilai k makin kecil pasti tidak memenuhi syarat x juga.

Jika k equals 0, maka x equals negative 37 comma 5 degree plus open parentheses 0 close parentheses times 90 degree equals negative 37 comma 5 degree plus 0 degree equals negative 37 comma 5 degree. Nilai tersebut memenuhi syarat x.

Jika k equals 1, maka x equals negative 37 comma 5 degree plus open parentheses 1 close parentheses times 90 degree equals negative 37 comma 5 degree plus 90 degree equals 52 comma 5 degree. Nilai tersebut memenuhi syarat x.

Jika k equals 2, maka x equals negative 37 comma 5 degree plus open parentheses 2 close parentheses times 90 degree equals negative 37 comma 5 degree plus 180 degree equals 142 comma 5 degree. Nilai tersebut tidak memenuhi syarat x sehingga untuk nilai k makin besar pasti tidak memenuhi syarat x juga.

Dengan demikian, diperoleh himpunan nilai x yang memenuhi yaitu open curly brackets negative 67 comma 5 degree comma space minus 37 comma 5 degree comma space 22 comma 5 degree comma space 52 comma 5 degree close curly brackets.

Jadi, jawaban yang tepat adalah A.

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