Pertanyaan

Selesaikanlah tiap persamaan berikut untuk 0 ≤ x ≤ 2 π . j. cos ( 2 x − 3 π ) sin 2 x = 2 1 3

Selesaikanlah tiap persamaan berikut untuk $0 \leq x \leq 2 π$ .

j. $cos (2 x - \frac{π}{3}) sin 2 x = \frac{1}{2} 3$

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S. Yoga

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Jawaban terverifikasi

Jawaban

himpunan penyelesaiannya adalah .

himpunan penyelesaiannya adalah $HP equals open curly brackets pi over 6 comma space pi over 4 comma thin space fraction numerator 2 pi over denominator 3 end fraction comma thin space fraction numerator 3 pi over denominator 4 end fraction comma thin space fraction numerator 7 pi over denominator 6 end fraction comma thin space fraction numerator 5 pi over denominator 4 end fraction comma thin space fraction numerator 5 pi over denominator 3 end fraction comma thin space fraction numerator 7 pi over denominator 4 end fraction close curly brackets$ .

Pembahasan

Ingat bahwa: Penyelesaiannya adalah sebagai berikut: Dari penyelesaian tersebut, yang memenuhi adalah Jadi, himpunan penyelesaiannya adalah .

Ingat bahwa:

Penyelesaiannya adalah sebagai berikut:

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses 2 x minus straight pi over 3 close parentheses space sin space 2 x end cell equals cell 1 half square root of 3 end cell row cell 1 half sin space open parentheses 2 x minus straight pi over 3 plus 2 x close parentheses minus 1 half sin space open parentheses 2 x minus straight pi over 3 minus 2 x close parentheses end cell equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell row cell 1 half sin space open parentheses 4 x minus straight pi over 3 close parentheses minus 1 half sin space open parentheses straight pi over 3 close parentheses end cell equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell row cell 1 half sin space open parentheses 4 x minus straight pi over 3 close parentheses minus 1 half open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses end cell equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell row cell 1 half sin space open parentheses 4 x minus straight pi over 3 close parentheses minus open parentheses fraction numerator square root of 3 over denominator 4 end fraction close parentheses end cell equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell row cell 1 half sin space open parentheses 4 x minus straight pi over 3 close parentheses end cell equals cell fraction numerator square root of 3 over denominator 4 end fraction end cell row cell sin space open parentheses 4 x minus straight pi over 3 close parentheses end cell equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards arrow cell sin open parentheses 4 x minus pi over 3 close parentheses equals fraction numerator square root of 3 over denominator 2 end fraction end cell row cell 4 x minus pi over 3 end cell equals cell pi over 3 plus 2 pi n left right double arrow x equals pi over 6 plus fraction numerator pi n over denominator 2 end fraction end cell row cell 4 x minus pi over 3 end cell equals cell fraction numerator 2 pi over denominator 3 end fraction plus 2 pi n left right double arrow x equals fraction numerator pi n over denominator 2 end fraction plus pi over 4 end cell end table$

Dari penyelesaian tersebut, yang memenuhi 0 less or equal than x less or equal than 2 straight pi adalah

$HP equals open curly brackets pi over 6 comma space pi over 4 comma thin space fraction numerator 2 pi over denominator 3 end fraction comma thin space fraction numerator 3 pi over denominator 4 end fraction comma thin space fraction numerator 7 pi over denominator 6 end fraction comma thin space fraction numerator 5 pi over denominator 4 end fraction comma thin space fraction numerator 5 pi over denominator 3 end fraction comma thin space fraction numerator 7 pi over denominator 4 end fraction close curly brackets$

Jadi, himpunan penyelesaiannya adalah $HP equals open curly brackets pi over 6 comma space pi over 4 comma thin space fraction numerator 2 pi over denominator 3 end fraction comma thin space fraction numerator 3 pi over denominator 4 end fraction comma thin space fraction numerator 7 pi over denominator 6 end fraction comma thin space fraction numerator 5 pi over denominator 4 end fraction comma thin space fraction numerator 5 pi over denominator 3 end fraction comma thin space fraction numerator 7 pi over denominator 4 end fraction close curly brackets$ .