The number of solutions of the equation sin t | vedclass

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Question

(B) Given equation: $\sin 	heta + \cos 	heta = \sin 2	heta$. Let $t = \sin 	heta + \cos 	heta$. Then $t^2 = \sin^2 	heta + \cos^2 	heta + 2 \si

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given equation: $\sin 	heta + \cos 	heta = \sin 2	heta$. Let $t = \sin 	heta + \cos 	heta$. Then $t^2 = \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = 1 + \sin 2	heta$. So,$\sin 2	heta = t^2 - 1$. The equation becomes $t = t^2 - 1$,or $t^2 - t - 1 = 0$. Solving for $t$,we get $t = \frac{1 \pm \sqrt{5}}{2}$. Since $t = \sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \frac{\pi}{4})$,the range of $t$ is $[-\sqrt{2}, \sqrt{2}]$. Note that $\sqrt{2} \approx 1.414$ and $\frac{1 + \sqrt{5}}{2} \approx 1.618$. Since $1.618 > 1.414$,the value $t = \frac{1 + \sqrt{5}}{2}$ is rejected. Thus,we must have $\sqrt{2} \sin(	heta + \frac{\pi}{4}) = \frac{1 - \sqrt{5}}{2}$. $\sin(	heta + \frac{\pi}{4}) = \frac{1 - \sqrt{5}}{2\sqrt{2}} \approx \frac{-1.236}{2.828} \approx -0.437$. Since $-1 In the interval $[-\pi, \pi]$,the shifted interval for $	heta + \frac{\pi}{4}$ is $[-\frac{3\pi}{4}, \frac{5\pi}{4}]$,which has length $2\pi$. Therefore,there are $2$ solutions.

The number of solutions of the equation $\sin \theta + \cos \theta = \sin 2\theta$ in the interval $[-\pi, \pi]$ is

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