Задание №6. Решить уравнение:

Question

Accepted Answer

№ варианта	a	б	в
1	\( 2\sin\left(x+\frac{\pi}{3}\right)=1 \)	\( 6\cos^2x - 7\cos x - 5 = 0 \)	\( \cos 2x = \sin\left(x+\frac{\pi}{2}\right) \)
2	\( 2\cos 2x = \sqrt{2} \)	\( \operatorname{tg}^2x + 5\operatorname{tg}x + 6 = 0 \)	\( \cos 2x + \sin^2x = 0,5 \)
3	\( 2\sin\left(x+\frac{\pi}{2}\right)=-\sqrt{2} \)	\( 2\cos^2x - 5\cos x + 2 = 0 \)	\( \cos 2x - 3\cos x + 2 = 0 \)
4	\( \sin\left(\frac{\pi}{2}-x\right)=\frac{\sqrt{2}}{2} \)	\( 8\sin^2x - 6\sin x - 5 = 0 \)	\( \cos 2x + \sin^2x = 0,75 \)
5	\( 2\sin 3x = \sqrt{3} \)	\( \sin^2x - 3\sin x + 2 = 0 \)	\( 2\cos^2x + 2\sin 2x - 3 \)
6	\( 2\cos\frac{x}{4} = -\sqrt{3} \)	\( 3\operatorname{tg}^2x + \operatorname{tg}x - 4 = 0 \)	\( \sqrt{3} \sin 2x + 3\cos 2x = 0 \)
7	\( 2\cos\left(x+\frac{\pi}{3}\right)=\sqrt{3} \)	\( \operatorname{tg}^2x - 2\operatorname{tg}x - 3 = 0 \)	\( \sin\left(\frac{7\pi}{2}+x\right) + 2\cos 2x = 1 \)
8	\( 2\operatorname{tg}\left(x+\frac{\pi}{4}\right)=2 \)	\( 2\sin^2x - 5\sin x + 2 = 0 \)	\( 6\sin^2x + 5\sin\left(\frac{\pi}{2}-x\right) - 2 = 0 \)
9	\( \operatorname{ctg}\left(x-\frac{\pi}{4}\right)=\sqrt{3} \)	\( 2\cos^2x + \cos x - 1 = 0 \)	\( 2\cos^2\left(\frac{3\pi}{2}+x\right) - \sin 2x \)
10	\( 2\cos\left(x-\frac{\pi}{3}\right)=\sqrt{3} \)	\( 2\sin^2x - 5\sin x + 2 = 0 \)	\( \cos 2x + \sin\left(\frac{\pi}{2}+x\right) + 1 = 0 \)

Ответ: