Trigonometric equations

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The number of solutions of $\sin^{2} x \cos^{2} x = 1 + \cos^{2} x \sin^{4} x$ in $[0, 2 π]$ is

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Solution

$W e h a v e \sin^{2} x \cos^{2} x = \cos^{2} x \sin^{4} x + 1$
$\Rightarrow \sin^{2} x \cos^{2} x - \cos^{2} x \sin^{4} x = 1$
$\begin{array}{l} \Rightarrow \sin^{2} x \cos^{2} x - \cos^{2} x \sin^{2} x \sin^{2} x = 1 \\ \Rightarrow \sin^{2} x \cos^{2} x (1 - \sin^{2} x) = 1 \\ \Rightarrow \sin^{2} x \cos^{4} x = 1 \end{array}$
No value of x can satisfy the above equation

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Trigonometric equations

Remember concepts with our Masterclasses.

The number of solutions of sin2xcos2x=1+cos2xsin4x in [0,2π] is

We have sin2xcos2x=cos2xsin4x+1⇒sin2xcos2x−cos2xsin4x=1⇒sin2xcos2x−cos2xsin2xsin2x=1⇒sin2xcos2x(1−sin2x)=1⇒sin2xcos4x=1No value of x can satisfy the above equation

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The number of solutions of $\sin^{2} x \cos^{2} x = 1 + \cos^{2} x \sin^{4} x$ in $[0, 2 π]$ is