Pertanyaan

Jika x 1 dan x 2 adalah solusi dari cos x sin 2 x 2 sin x cos 2 x + 3 tan x + 27 = 0 , maka adalah ....

Jika $x_{1}$ dan $x_{2}$ adalah solusi dari $\frac{2 sin x cos 2 x}{cos x sin 2 x} + 3 tan x + 27 = 0$ , maka $begin mathsize 14px style fraction numerator tan invisible function application open parentheses x subscript 1 plus x subscript 2 close parentheses over denominator tan invisible function application open parentheses x subscript 1 minus x subscript 2 close parentheses end fraction end style$ adalah ....

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

Pembahasan

Perhatikan bahwa Karena , maka Perhatikan bahwa dan Sehingga Jadi, jawaban yang tepat adalah A.

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 sin invisible function application x cos invisible function application 2 x over denominator cos invisible function application x sin invisible function application 2 x end fraction plus 3 tan invisible function application x plus 27 end cell equals 0 row cell 2 times fraction numerator sin invisible function application x over denominator cos invisible function application x end fraction times fraction numerator cos invisible function application 2 x over denominator sin invisible function application 2 x end fraction plus 3 tan invisible function application x plus 27 end cell equals 0 row cell 2 times tan invisible function application x times fraction numerator 1 over denominator tan invisible function application 2 x end fraction plus 3 tan invisible function application x plus 27 end cell equals 0 row cell fraction numerator 2 tan invisible function application x over denominator tan invisible function application 2 x end fraction plus 3 tan invisible function application x plus 27 end cell equals 0 end table end style$

Karena $begin mathsize 14px style tan invisible function application 2 x equals fraction numerator 2 tan invisible function application x over denominator 1 minus tan squared invisible function application x end fraction end style$ , maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 tan invisible function application x over denominator tan invisible function application 2 x end fraction plus 3 tan invisible function application x plus 27 end cell equals 0 row cell fraction numerator 2 tan invisible function application x over denominator fraction numerator 2 tan invisible function application x over denominator 1 minus tan squared invisible function application x end fraction end fraction plus 3 tan invisible function application x plus 27 end cell equals 0 row cell 1 minus tan squared invisible function application x plus 3 tan invisible function application x plus 27 end cell equals 0 row cell negative tan squared invisible function application x plus 3 tan invisible function application x plus 28 end cell equals 0 row cell tan squared invisible function application x minus 3 tan invisible function application x minus 28 end cell equals 0 row cell open parentheses tan invisible function application x plus 4 close parentheses open parentheses tan invisible function application x minus 7 close parentheses end cell equals 0 row cell tan invisible function application x subscript 1 end cell equals cell negative 4 text atau end text tan invisible function application x subscript 2 equals 7 end cell end table end style$

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan invisible function application open parentheses x subscript 1 plus x subscript 2 close parentheses end cell equals cell fraction numerator tan invisible function application x subscript 1 plus tan invisible function application x subscript 2 over denominator 1 minus tan invisible function application x subscript 1 tan invisible function application x subscript 2 end fraction end cell row blank equals cell fraction numerator negative 4 plus 7 over denominator 1 minus open parentheses negative 4 close parentheses open parentheses 7 close parentheses end fraction end cell row blank equals cell fraction numerator 3 over denominator 1 plus 28 end fraction end cell row blank equals cell 3 over 29 end cell end table end style$

dan

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan invisible function application open parentheses x subscript 1 minus x subscript 2 close parentheses end cell equals cell fraction numerator tan invisible function application x subscript 1 minus tan invisible function application x subscript 2 over denominator 1 plus tan invisible function application x subscript 1 tan invisible function application x subscript 2 end fraction end cell row blank equals cell fraction numerator negative 4 minus 7 over denominator 1 plus open parentheses negative 4 close parentheses times 7 end fraction end cell row blank equals cell fraction numerator negative 11 over denominator 1 minus 28 end fraction end cell row blank equals cell fraction numerator negative 11 over denominator negative 27 end fraction end cell row blank equals cell 11 over 27 end cell end table end style$

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan invisible function application open parentheses x subscript 1 plus x subscript 2 close parentheses over denominator tan invisible function application open parentheses x subscript 1 minus x subscript 2 close parentheses end fraction end cell equals cell fraction numerator 3 over 29 over denominator 11 over 27 end fraction end cell row blank equals cell 3 over 29 cross times 27 over 11 end cell row blank equals cell 81 over 319 end cell end table end style$

Jadi, jawaban yang tepat adalah A.