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If $5\left ( \tan ^{2}x-\cos ^{2}x \right )=2\cos \: 2x+9$

then the value of $\cos 4x$ is :

Option 1)
$\frac{1}{3}$

Option 2)
$\frac{2}{9}$

Option 3)
$-\frac{7}{9}$

Option 4)
$-\frac{3}{5}$

Answers (1)

As we learnt in

Trigonometric Equations -

The equations involving trigonometric function of unknown angles are known as trigonometric equations.

- wherein

e.g. $\cos ^{2}\Theta - 4\cos \Theta = 1$

$5 (\tan^{2}x-\cos^{2}x)=2\cos 2x+9$ ...............( 1 )

Now, $\tan^{2}x=\frac{1-\cos2x}{1+\cos2x}$ and, $\cos^{2}x=\frac{1}{2}[1+\cos2x]$

Let $\cos2x=t$

Then, ( 1 ) becomes:

$5\left ( \frac{1-t}{1+t} \right )-\frac{5}{2}(1+t)=2t+9$

$\Rightarrow 5-5t-\frac{5}{2}(1+t^{2}+2t)=(2t+9)(t+1)$

$\Rightarrow \frac{-5}{2}t^{2}-10t+\frac{5}{2}=2t^{2}+11t+9$

$\Rightarrow \frac{9}{2}t^{2}+21t+\frac{13}{2}=0$

$\Rightarrow9t^{2}+42t+13=0$

$t=\frac{-1}{3}$ $[\left| t\right|>1\:\:\: neglected]$

$\Rightarrow \cos 2x=\frac{-1}{3}$

$\Rightarrow \cos^{2} 2x-1=\frac{2}{9}-1=-\frac{7}{9}$

$\Rightarrow \cos 4x=-\frac{7}{9}$

Option 1)

$\frac{1}{3}$

This option is incorrect.

Option 2)

$\frac{2}{9}$

This option is incorrect.

Option 3)

$-\frac{7}{9}$

This option is correct.

Option 4)

$-\frac{3}{5}$

This option is incorrect.

Posted by

Sabhrant Ambastha

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