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Help me answer: - Trigonometry - JEE Main-7

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)
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V Vakul

 

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

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