Pertanyaan

Himpunan penyelesaian persamaan cos 2 x − sin x = 0 , untuk 0 ≤ x ≤ 2 π adalah ....

Himpunan penyelesaian persamaan $cos 2 x - sin x = 0$ , untuk $0 \leq x \leq 2 π$ adalah ....

$begin mathsize 14px style open curly brackets straight pi over 6 comma fraction numerator 5 straight pi over denominator 6 end fraction comma fraction numerator 3 straight pi over denominator 2 end fraction close curly brackets end style$
$begin mathsize 14px style open curly brackets straight pi over 2 comma fraction numerator 3 straight pi over denominator 2 end fraction comma fraction numerator 7 straight pi over denominator 6 end fraction close curly brackets end style$
$begin mathsize 14px style open curly brackets fraction numerator 7 straight pi over denominator 6 end fraction comma fraction numerator 4 straight pi over denominator 3 end fraction comma fraction numerator 11 straight pi over denominator 6 end fraction close curly brackets end style$
$begin mathsize 14px style open curly brackets fraction numerator 4 straight pi over denominator 3 end fraction comma fraction numerator 11 straight pi over denominator 6 end fraction comma 2 straight pi close curly brackets end style$

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Klaim

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

INGAT Rumus sudut rangkap atau Persamaan trigonometri pada sin Menentukan nilai pada Menentukan nilai pada Himpunan penyelesaian adalah Jadi, jawaban yang tepat adalah B.

INGAT Rumus sudut rangkap

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 2 a end cell equals cell 1 minus 2 space sin squared space a end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space sin space x minus 1 end cell equals cell 0 space end cell row cell sin space x end cell equals cell 1 half end cell row cell sin space x space end cell equals cell sin space straight pi over 6 space space space space space end cell end table atau $table attributes columnalign right center left columnspacing 0px end attributes row cell sin space x space plus 1 end cell equals 0 row cell sin space x space end cell equals cell space minus 1 end cell row cell sin space x space end cell equals cell space sin fraction numerator 3 straight pi over denominator 2 end fraction end cell end table$

Persamaan trigonometri pada sin

Menentukan nilai x subscript 1 space dan space x subscript 2 pada sin space x space equals sin space straight pi over 6

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell straight pi over 6 plus k times 2 straight pi end cell row bold k bold equals bold 0 row cell straight x subscript 1 end cell equals cell straight pi over 6 plus 0 times 2 straight pi end cell row cell straight x subscript 1 end cell equals cell straight pi over 6 end cell row blank blank blank row blank blank blank row cell straight x subscript 2 end cell equals cell open parentheses straight pi minus straight pi over 6 close parentheses plus k times 2 straight pi end cell row cell straight x subscript 2 end cell equals cell fraction numerator 5 straight pi over denominator 6 end fraction plus k italic times 2 straight pi end cell row bold k bold equals bold 0 row cell straight x subscript 2 end cell equals cell fraction numerator 5 straight pi over denominator 6 end fraction plus 0 times 2 straight pi end cell row cell straight x subscript 2 end cell equals cell fraction numerator 5 straight pi over denominator 6 end fraction end cell end table$

Menentukan nilai x subscript 1 space dan space x subscript 2 pada $sin space x space equals sin space fraction numerator 3 straight pi over denominator 2 end fraction$

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell fraction numerator 3 straight pi over denominator 2 end fraction plus k times 2 straight pi end cell row bold k bold equals bold 0 row cell straight x subscript 1 end cell equals cell fraction numerator 3 straight pi over denominator 2 end fraction plus 0 times 2 straight pi end cell row cell straight x subscript 1 end cell equals cell fraction numerator 3 straight pi over denominator 2 end fraction end cell row blank blank blank row blank blank blank row cell straight x subscript 2 end cell equals cell open parentheses straight pi minus fraction numerator 3 straight pi over denominator 2 end fraction close parentheses plus k times 2 straight pi end cell row cell straight x subscript 2 end cell equals cell negative straight pi over 2 plus k italic times 2 straight pi end cell row bold k bold equals bold 1 row cell straight x subscript 2 end cell equals cell negative straight pi over 2 plus 1 times 2 straight pi end cell row cell straight x subscript 2 end cell equals cell fraction numerator 3 straight pi over denominator 2 end fraction end cell end table$

Himpunan penyelesaian adalah $open curly brackets straight pi over 6 comma fraction numerator 5 straight pi over denominator 6 end fraction comma fraction numerator 3 straight pi over denominator 2 end fraction close curly brackets$

Jadi, jawaban yang tepat adalah B.