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The number of solutions of the equation $1+\sin ^{4}x=\cos ^{2}3x,$ $x\equiv \left [ -\frac{5\pi }{2},\frac{5\pi }{2} \right ]$ is :

Option 1)
$4$

Option 2)
$3$

Option 3)
$5$

Option 4)
$7$

Answers (1)

Trigonometric Identities -

$\sin ^{2}\Theta + \cos ^{2}\Theta = 1$

$1 + \tan ^{2}\Theta = \sec ^{2}\Theta$

$1 + \cot ^{2}\Theta = cosec ^{2}\Theta$

- wherein

They are true for all real values of $\Theta$

Results from General Solution -

$\sin \Theta = 0 \Rightarrow \Theta = n\pi$

-

$1+\sin ^{4}x=\cos ^{2}3x$ $x\equiv \left [ \frac{-5\pi }{2},\frac{5\pi }{2} \right ]$

Rewrite the trignometric equation

$1-\cos ^{2}\left ( 3x \right )+\sin ^{4}x=0$

$\because \sin ^{2}\left ( x \right )+\cos ^{2}\left ( x \right )=1$

$\sin ^{2}\left ( 3x \right )+\sin ^{4}\left ( x \right )=0$

$\Rightarrow \sin \left ( 3x \right )=0\: \: \: \: \: \: and\: \: \: \: \sin \left ( x \right )=0$

$\sin \left ( x \right )=0$ at $x=0,\pi ,-\pi,-2\pi,2\pi$

This value is also true for $\sin \left ( 3x \right )$

Option 1)

$4$

Option 2)

$3$

Option 3)

$5$

Option 4)

$7$

Posted by

Vakul

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