Question

СРОЧНО ПОМОГИТЕ РЕШИТЬ!!!
cos3x + sin2x - cos x =0

Answer 1

$cos 3x + sin 2x - cos x = 0$

$cos 3x - cos x + sin 2x = 0$

По формуле $cos alpha - cos beta = -2sin dfrac{alpha - beta } {2} sin dfrac{alpha + beta } {2}$ имеем:

$-2sin x sin 2x + sin 2x = 0$

$sin2x (-2sin x + 1) = 0$

$left[begin{array}{ccc}sin 2x = 0 \-2sin x + 1 = 0\end{array}right$

$left[begin{array}{ccc} 2x = pi k, k in Z \sin x = dfrac{1}{2} \end{array}right$

$left[begin{array}{ccc} x = dfrac{pi k}{2} , k in Z \ x = (-1)^{n}dfrac{pi}{6} + pi n, n in Z \end{array}right$

Ответ: $dfrac{pi k}{2}; (-1)^{n}dfrac{pi}{6} + pi n; k, n in Z$