The number of solution(s) of the equation sin | vedclass

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Question

(B) Given equation: $\sin 2	heta + \cos 2	heta = -\frac{1}{2}$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin 2	heta + \frac{1}{\sqrt{2}}

Vedclass · Accepted Answer

$1$

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(B) Given equation: $\sin 2	heta + \cos 2	heta = -\frac{1}{2}$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin 2	heta + \frac{1}{\sqrt{2}} \cos 2	heta = -\frac{1}{2\sqrt{2}}$$\sin(2	heta + \frac{\pi}{4}) = -\frac{1}{2\sqrt{2}}$Since $	heta \in (0, \frac{\pi}{2})$,we have $2	heta \in (0, \pi)$,so $2	heta + \frac{\pi}{4} \in (\frac{\pi}{4}, \frac{5\pi}{4})$.Let $\alpha = 2	heta + \frac{\pi}{4}$. We need to find the number of solutions for $\sin \alpha = -\frac{1}{2\sqrt{2}}$ in the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$.Since $\sin \alpha$ is negative in the third and fourth quadrants,and our interval $(\frac{\pi}{4}, \frac{5\pi}{4})$ covers the third quadrant,there is exactly one value of $\alpha$ in $(\pi, \frac{5\pi}{4})$ such that $\sin \alpha = -\frac{1}{2\sqrt{2}}$.Thus,there is exactly $1$ solution.

The number of solution$(s)$ of the equation $\sin 2\theta + \cos 2\theta = -\frac{1}{2}$ for $\theta \in \left( 0, \frac{\pi}{2} \right)$ is:

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