How to solve this trigonometric equation?

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Answer 1 · 2018-04-01T13:13:58

Answer is (3).

$5 ({tan}^{2} x - {cos}^{2} x) = 2 cos 2 x + 9$ $5(tan^2x-cos^2x)=2cos2x+9$

$\Leftrightarrow 5 ({sec}^{2} x - 1) - 5 {cos}^{2} x = 2 (2 {cos}^{2} x - 1) + 9$ $hArr5(sec^2x-1)-5cos^2x=2(2cos^2x-1)+9$

or $5 {sec}^{2} x - 9 {cos}^{2} x - 12 = 0$ $5sec^2x-9cos^2x-12=0$ and multiplying by ${cos}^{2} x$ $cos^2x$ we get

$5 - 9 {cos}^{4} x - 12 {cos}^{2} x = 0$ $5-9cos^4x-12cos^2x=0$

i.e. $9 {cos}^{4} x + 12 {cos}^{2} x - 5 = 0$ $9cos^4x+12cos^2x-5=0$

and ${cos}^{2} x = \frac{- 12 \pm \sqrt{144 + 180}}{18}$ $cos^2x=(-12+-sqrt(144+180))/18$

= $\frac{- 12 \pm 18}{18}$ $(-12+-18)/18$

i.e. ${cos}^{2} x = \frac{1}{3}$ $cos^2x=1/3$ as we cannot have ${cos}^{2} x$ $cos^2x$ as negative

Hence $cos 2 x = 2 {cos}^{2} x - 1 = 2 \cdot \frac{1}{3} - 1 = - \frac{1}{3}$ $cos2x=2cos^2x-1=2*1/3-1=-1/3$

and $cos 4 x = 2 {cos}^{2} 2 x - 1 = 2 {(- \frac{1}{3})}^{2} - 1 = - \frac{7}{9}$ $cos4x=2cos^2 2x-1=2(-1/3)^2-1=-7/9$

Hence, answer is (3).