Total number of solutions of sin4⁡x+cos4⁡x=sin⁡x⋅cos⁡x in [0,2π] is equal to

Solution:

Given, $\sin^{4} x + \cos^{4} x = \sin x \cdot \cos x$

$\begin{array}{ll} \Rightarrow & {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cdot \cos^{2} x = \sin x \cdot \cos x \\ \Rightarrow & 1 - \frac{\sin^{2} 2 x}{2} = \frac{\sin 2 x}{2} \\ \Rightarrow & \sin^{2} 2 x + \sin 2 x - 2 = 0 \\ \Rightarrow & (\sin 2 x + 2) (\sin 2 x - 1) = 0 \\ \Rightarrow & \sin 2 x = 1 \\ ∴ & 2 x = (4 n + 1) \frac{π}{2} \\ \Rightarrow & x = (4 n + 1) \frac{π}{4} \Rightarrow x = \frac{π}{4}, \frac{5 π}{4} \end{array}$

Hence, two solutions exist.

Total number of solutions of $\sin^{4} x + \cos^{4} x = \sin x \cdot \cos x in [0, 2 π]$ is equal to

Solution:

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