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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)


  • Option 2)


  • Option 3)


  • Option 4)



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


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)


This option is incorrect.

Option 2)


This option is incorrect.

Option 3)


This option is correct.

Option 4)


This option is incorrect.

Posted by

Sabhrant Ambastha

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