The common solution set of the equations 2 si | vedclass

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(A) Given equations are:$2 \sin^2 x + \sin^2 2x = 2$  ... $(i)$$\sin 2x + \cos 2x = 	an x$  ... (ii)From $(i)$:$2 \sin^2 x + (2 \sin x \cos x)^2 = 2$

Vedclass · Accepted Answer

$\left\{x \in R : x = (2n + 1) \frac{\pi}{4}, n \in Z\right\}$

Vedclass · Answer

(A) Given equations are:$2 \sin^2 x + \sin^2 2x = 2$  ... $(i)$$\sin 2x + \cos 2x = 	an x$  ... (ii)From $(i)$:$2 \sin^2 x + (2 \sin x \cos x)^2 = 2$$2 \sin^2 x + 4 \sin^2 x \cos^2 x = 2$$2 \sin^2 x (1 + 2 \cos^2 x) = 2$$2 \sin^2 x (1 + 2(1 - \sin^2 x)) = 2$$2 \sin^2 x (3 - 2 \sin^2 x) = 2$$6 \sin^2 x - 4 \sin^4 x = 2$$2 \sin^4 x - 3 \sin^2 x + 1 = 0$$(2 \sin^2 x - 1)(\sin^2 x - 1) = 0$$\sin^2 x = \frac{1}{2}$ or $\sin^2 x = 1$If $\sin^2 x = \frac{1}{2}$,then $\cos 2x = 1 - 2 \sin^2 x = 0$,so $2x = (2k+1)\frac{\pi}{2} \Rightarrow x = (2k+1)\frac{\pi}{4}$.If $\sin^2 x = 1$,then $\cos^2 x = 0$,so $x = (2k+1)\frac{\pi}{2}$.From (ii):$\frac{2 	an x}{1 + 	an^2 x} + \frac{1 - 	an^2 x}{1 + 	an^2 x} = 	an x$$2 	an x + 1 - 	an^2 x = 	an x + 	an^3 x$$	an^3 x + 	an^2 x - 	an x - 1 = 0$$	an^2 x (	an x + 1) - 1 (	an x + 1) = 0$$(	an^2 x - 1)(	an x + 1) = 0$$(	an x - 1)(	an x + 1)^2 = 0$$	an x = 1$ or $	an x = -1$$x = n\pi + \frac{\pi}{4}$ or $x = n\pi - \frac{\pi}{4}$,which is $x = (2n \pm 1)\frac{\pi}{4}$.Comparing the solutions,the common set is $x = (2n + 1)\frac{\pi}{4}$.

The common solution set of the equations $2 \sin^2 x + \sin^2 2x = 2$ and $\sin 2x + \cos 2x = \tan x$ is

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