Question

Даю 100 баллов, помогите решить

Answer 1

Ответ:

4.

а) $\boxed{\dfrac{1 - \cos 4\alpha }{\sin^{2} 2\alpha } = 2}$

б) $\boxed{\dfrac{\sin 2x - 2 \sin x}{\cos x - \sin ^{2} 3x - \cos^{2} 3x} = 2 \sin x}$

5.

а) $\boxed{x = б \dfrac{\arccos(\sqrt{2})}{8} + \dfrac{\pi n}{4} , n \in \mathbb Z}$

б) $\boxed{x = \dfrac{arctg(-1,5) }{\pi } + n, n \in \mathbb Z}$

Объяснение:

4.

а) $\dfrac{1 - \cos 4\alpha }{\sin^{2} 2\alpha } = \dfrac{1 -( \cos^{2} 2\alpha - \sin ^{2} 2\alpha ) }{\sin^{2} 2\alpha } = \dfrac{\cos^{2} 2\alpha + \sin ^{2} 2\alpha - \cos^{2} 2\alpha + \sin ^{2} 2\alpha }{\sin^{2} 2\alpha } =$

$= \dfrac{2 \sin ^{2} 2\alpha }{\sin^{2} 2\alpha } = 2$

б) $\dfrac{\sin 2x - 2 \sin x}{\cos x - \sin ^{2} 3x - \cos^{2} 3x} = \dfrac{2\sin x\cos x - 2 \sin x}{\cos x -( \sin ^{2} 3x + \cos^{2} 3x)} = \dfrac{2\sin x(\cos x - 1)}{\cos x -( 1)} =$

$= \dfrac{2\sin x(\cos x - 1)}{(\cos x - 1)} = 2 \sin x$

5.

а) $2 \cos^{2} 4x - 2 \sin^{2} 4x - 4 \cos 8x = -\sqrt{2}$

$2( \cos^{2} 4x - \sin^{2} 4x) - 4 \cos 8x = -\sqrt{2}$

$2 \cos 8x - 4 \cos 8x = -\sqrt{2}$

$-2 \cos 8x = -\sqrt{2}|\cdot(-1)$

$\cos 8x = \sqrt{2}$

$8x = б \arccos(\sqrt{2)} + 2\pi n, n \in \mathbb Z|:8$

$x = б \dfrac{\arccos(\sqrt{2})}{8} + \dfrac{2\pi n}{8} , n \in \mathbb Z$

$x = б \dfrac{\arccos(\sqrt{2})}{8} + \dfrac{\pi n}{4} , n \in \mathbb Z$

б) $3 \sin 2\pi x + 5\cos 2\pi x = - \sin^{2} \pi x - 4$

Замена: $\pi x = t$

$3 \sin 2t + 5\cos 2t = - \sin^{2} t - 4$

$3 \sin 2t + 5(1 - 2\sin^{2} t) + \sin^{2} t = - 4$

$3 \sin 2t + 5 - 10\sin^{2} t + \sin^{2} t = - 4$

$3 \sin 2t - 9\sin^{2} t = -9|:3$

$\sin 2t - 3\sin^{2} t = -3$

$2 \sin t \cos t - 3\sin^{2} t = -3$

$3\sin^{2} t - 2 \sin t \cos t - 3 = 0 |: \cos^{2}t$

$3 \ tg^{2} t - 2\ tg\ t - 3 \cdot ( 1 + tg^{2} t) = 0$

$3 \cdot tg^{2} t - 2\ tg\ t - 3 - 3\cdot tg^{2} t = 0$

$- 2\ tg\ t - 3 = 0$

$- 2\ tg\ t = 3|:(-2)$

$tg \ t = -1,5$

$t = arctg(-1,5) + \pi n, n \in \mathbb Z$

$\pi x = arctg(-1,5) + \pi n, n \in \mathbb Z|:\pi$

$x = \dfrac{arctg(-1,5) }{\pi } + n, n \in \mathbb Z$