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Q.
The total number of solution of $sin^4x + cos^4x = sinx\, cosx$
in $[0, 2\pi]$ is equal to ______
Trigonometric Functions
Solution:
$sin^4 x + cos^4 x = sin x\,cos x$
or $(sin^2 x + cos^2 x)^2 - 2\, sin^2 x \,cos^2 x = sin\,x\,cos\,x$
or $ 1 - \frac{sin^2\,2x}{2} = \frac{sin\,2x}{2}$
or $sin^2 2x + sin\,2x - 2 = 0$
or $(sin\,2x + 2)(sin \,2x - 1) = 0$
or $sin\,2x = 1$
or $2x = (4n + 1) \frac{\pi}{2}, n \in Z$
or$x = (4n + 1) \frac{\pi}{4}, n \in Z$
$= \frac{\pi}{4}, \frac{5\pi}{4} (\because x \in [0, 2\pi])$
Thus, there are two solutions.