The number of solutions for the equation sin | vedclass

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Question

(A) Given equation: $\sin 2x + \cos 4x = 2$ Using the identity $\cos 4x = 1 - 2\sin^2 2x$,we get: $\sin 2x + 1 - 2\sin^2 2x = 2$ Rearranging the terms

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(A) Given equation: $\sin 2x + \cos 4x = 2$ Using the identity $\cos 4x = 1 - 2\sin^2 2x$,we get: $\sin 2x + 1 - 2\sin^2 2x = 2$ Rearranging the terms: $2\sin^2 2x - \sin 2x + 1 = 0$ Let $t = \sin 2x$. The equation becomes $2t^2 - t + 1 = 0$. The discriminant $D$ of this quadratic equation is $D = (-1)^2 - 4(2)(1) = 1 - 8 = -7$. Since $D Therefore,there are no real solutions for $x$.

The number of solutions for the equation $\sin 2x + \cos 4x = 2$ is

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