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$\quad$ If $\cos 7 \theta=\cos \theta-\sin 4 \theta,$ then the general value
of $\theta$ is
Options:
of $\theta$ is
Solution:
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Verified Answer
The correct answer is:
$\frac{n \pi}{4}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$
$\cos 7 \theta=\cos \theta-\sin 4 \theta$
$\Rightarrow \sin 4 \theta=\cos \theta-\cos 7 \theta$
$\Rightarrow \sin 4 \theta=2 \sin 4 \theta \sin 3 \theta$
$\Rightarrow \sin 4 \theta(1-2 \sin 3 \theta)=0$
$\therefore \sin 4 \theta=0$ or $\sin 3 \theta=\frac{1}{2}$
$\Rightarrow 4 \theta=n \pi$ or $3 \theta=n \pi+(-1)^{n} \frac{\pi}{6}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $\frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$
$\Rightarrow \sin 4 \theta=\cos \theta-\cos 7 \theta$
$\Rightarrow \sin 4 \theta=2 \sin 4 \theta \sin 3 \theta$
$\Rightarrow \sin 4 \theta(1-2 \sin 3 \theta)=0$
$\therefore \sin 4 \theta=0$ or $\sin 3 \theta=\frac{1}{2}$
$\Rightarrow 4 \theta=n \pi$ or $3 \theta=n \pi+(-1)^{n} \frac{\pi}{6}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $\frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$
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