Pertanyaan

Tentukan himpunan penyelesaian dari 2 ⋅ cos ( 2 x + 2 π ) − 2 = 0 , dengan − π ≤ x ≤ π !

Tentukan himpunan penyelesaian dari $2 \cdot cos (2 x + \frac{π}{2}) - 2 = 0$ , dengan $- π \leq x \leq π$ !

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himpunan penyelesaiannya adalah .

himpunan penyelesaiannya adalah $open curly brackets negative fraction numerator 3 straight pi over denominator 8 end fraction comma negative straight pi over 8 comma fraction numerator 5 straight pi over denominator 8 end fraction comma fraction numerator 7 straight pi over denominator 8 end fraction close curly brackets$ .

Pembahasan

Persamaan trigonometri pada cos Menentukan nilai Menentukan nilai Jadi, himpunan penyelesaiannya adalah .

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 times cos left parenthesis 2 x plus straight pi over 2 right parenthesis minus square root of 2 end cell equals 0 row cell fraction numerator 2 times cos left parenthesis 2 x plus straight pi over 2 right parenthesis over denominator 2 end fraction end cell equals cell fraction numerator square root of 2 over denominator 2 end fraction end cell row cell cos left parenthesis 2 x plus straight pi over 2 right parenthesis end cell equals cell 1 half square root of 2 end cell row cell cos open parentheses 2 x plus straight pi over 2 close parentheses end cell equals cell cos space straight pi over 4 end cell end table$

Persamaan trigonometri pada cos

cos space x space equals space cos space alpha x subscript 1 equals alpha plus k.2 straight pi straight x subscript 2 equals negative straight alpha plus k times 2 straight pi

Menentukan nilai

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell alpha plus k times 2 straight pi end cell row cell 2 straight x plus straight pi over 2 end cell equals cell straight pi over 4 plus k times 2 straight pi end cell row cell 2 straight x end cell equals cell negative straight pi over 4 plus k times 2 straight pi end cell row cell straight x subscript 1 end cell equals cell negative straight pi over 8 plus k times straight pi end cell row blank blank blank row cell bold subtitusi bold space bold italic k end cell bold equals bold 0 row cell straight x subscript 1 end cell equals cell negative straight pi over 8 plus 0 times straight pi end cell row cell straight x subscript 1 end cell equals cell negative straight pi over 8 end cell row cell bold subtitusi bold space bold italic k end cell bold equals bold 1 row cell x subscript italic 1 end cell italic equals cell italic minus pi over italic 8 plus 1 times straight pi end cell row cell straight x subscript 1 end cell equals cell fraction numerator 7 straight pi over denominator 8 end fraction end cell end table$

Menentukan nilai

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell negative alpha plus k times 2 straight pi end cell row cell 2 straight x plus straight pi over 2 end cell equals cell negative straight pi over 4 plus k times 2 straight pi end cell row cell 2 straight x end cell equals cell negative fraction numerator 3 straight pi over denominator 4 end fraction plus k times 2 straight pi end cell row cell straight x subscript 1 end cell equals cell negative fraction numerator 3 straight pi over denominator 8 end fraction plus k times straight pi end cell row blank blank blank row cell bold subtitusi bold space bold italic k end cell bold equals bold 0 row cell straight x subscript 1 end cell equals cell negative fraction numerator 3 straight pi over denominator 8 end fraction plus 0 times straight pi end cell row cell straight x subscript 1 end cell equals cell negative fraction numerator 3 straight pi over denominator 8 end fraction end cell row cell bold subtitusi bold space bold italic k end cell bold equals bold 1 row cell x subscript italic 1 end cell italic equals cell italic minus fraction numerator 3 pi over denominator italic 8 end fraction plus 1 times straight pi end cell row cell straight x subscript 1 end cell equals cell fraction numerator 5 straight pi over denominator 8 end fraction end cell end table$

Jadi, himpunan penyelesaiannya adalah $open curly brackets negative fraction numerator 3 straight pi over denominator 8 end fraction comma negative straight pi over 8 comma fraction numerator 5 straight pi over denominator 8 end fraction comma fraction numerator 7 straight pi over denominator 8 end fraction close curly brackets$ .

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