Pertanyaan

Jika x 1 dan x 2 adalah solusi dari cos x ⋅ sin 2 x 2 sin x cos 2 x − 5 tan x + 5 = 0 maka

Jika $x_{1}$ dan $x_{2}$ adalah solusi dari $\frac{2 sin x cos 2 x}{cos x \cdot sin 2 x} - 5 tan x + 5 = 0$ maka

$begin mathsize 14px style fraction numerator square root of 5 over denominator 7 end fraction end style$
$begin mathsize 14px style fraction numerator square root of 5 over denominator 3 end fraction end style$

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Jawaban terverifikasi

Pembahasan

Sederhanakan bentuk trigonometri di bawah ini terlebih dahulu. Maka,

Sederhanakan bentuk trigonometri di bawah ini terlebih dahulu.

$begin mathsize 14px style fraction numerator 2 space sin space straight x space cos space 2 straight x over denominator cos space straight x space sin space 2 straight x end fraction minus 5 space tan space straight x plus 5 equals 0 fraction numerator 2 fraction numerator sin space straight x over denominator cos space straight x end fraction over denominator fraction numerator sin space 2 straight x over denominator cos space 2 straight x end fraction end fraction minus 5 space tan space straight x plus 5 equals 0 fraction numerator 2 space tan space straight x over denominator tan space 2 straight x end fraction minus 5 space tan space straight x plus 5 equals 0 fraction numerator 2 space tan space straight x over denominator fraction numerator 2 space tan space straight x over denominator 1 minus tan squared space straight x end fraction end fraction minus 5 space tan space straight x plus 5 equals 0 1 minus tan squared space straight x minus 5 space tan space straight x plus 5 equals 0 tan squared space straight x plus 5 space tan space straight x minus 6 equals 0 open parentheses tan space straight x plus 6 close parentheses open parentheses tan space straight x minus 1 close parentheses equals 0 tan space straight x equals negative 6 space atau space tan space straight x equals 1 end style$

Maka,

$begin mathsize 14px style tan space open parentheses straight x subscript 1 plus straight x subscript 2 close parentheses equals fraction numerator tan space straight x subscript 1 plus tan space straight x subscript 2 over denominator 1 minus tanx subscript 1 tanx subscript 2 end fraction fraction numerator negative 6 plus 1 over denominator 1 minus open parentheses negative 6 close parentheses open parentheses 1 close parentheses end fraction equals negative 5 over 7 end style$