Question

81. If 5(tan2x - cos2x) = 2cos 2x + 9, then the value of cos4x is :-​

Answer 1

Answer : 7/9 hope itz helpfull to you. drop some thanks

Answer 2

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : value \: of \: cos4x \\ \\ given \: that \\ 5(tan {}^{2} x - cos {}^{2}x) = 2cos2x + 9 \\ \\ 5[( sec {}^{2} x - 1 - cos {}^{2} x) = 2(2cos {}^{2} x - 1) + 9 \\ \\ 5[( \frac{1}{cos {}^{2} x} - cos {}^{2} x - 1)] = 4cos {}^{2} x - 2 + 9 \\ \\ 5[( \frac{1 - cos {}^{4} x - cos {}^{2}x }{cos {}^{2} x} )] = 4cos {}^{2} x + 7 \\ \\ 5 - 5cos {}^{4} x - 5cos {}^{2} x = 4.cos {}^{4} x + 7cos {}^{2} x \\ \\ 9cos {}^{4} x + 12cos {}^{2} x - 5 = 0 \\ \\ 9cos {}^{4} x + 15cos {}^{2} x - 3cos {}^{2} x - 5 = 0 \\ \\ 3cos {}^{2} x(3cos {}^{2} x + 5) - 1(3cos {}^{2} x + 5) = 0 \\ \\ (3cos {}^{2} x + 5)(3cos {}^{2} x - 1) = 0 \\ \\ cos {}^{2} x = - \frac{5}{3} \: \: or \: \: cos {}^{2} x = \frac{1}{3} \\ \\ but \: since \: we \: know \: that \\ cosine \: function \: ∈ \: ( - 1,0,1) \\ \\ cos {}^{2} x = - \frac{5}{3} \: \: is \: absurd$

$so \: we \: know \: that \: \\ cos4x = 2cos {}^{2} 2x - 1 \\ = 2.(cos2x ){}^{2} - 1 \\ = 2.(2cos {}^{2} x - 1) {}^{2} - 1 \\ \\ = 2(2 \times \frac{1}{3} - 1) {}^{2} - 1 \\ \\ = 2( \frac{2 - 3}{3} ) {}^{2} - 1 \\ \\ = 2( \frac{ - 1}{3} ) {}^{2} - 1 \\ \\ = \frac{2}{9} - 1 \\ \\ = - \frac{7}{9}$