Question

Кому не трудно, помогите пожалуйста

Answer 1

$\displaystyle \frac{sin^{2}(x)}{2} =\frac{1}{3}$

$\displaystyle \frac{1-cos^{2}(x)}{2} =\frac{1}{3}$

$\displaystyle \frac{1}{2}-\frac{cos^{2}(x)}{2} =\frac{1}{3}$

$\displaystyle \frac{1}{2} -\frac{1}{3} =\frac{cos^{2}(x)}{2}$

$\displaystyle \frac{cos^{2}(x)}{2} =\frac{1}{6}$

$\displaystyle 6cos^{2}(x) = 2$

$\displaystyle cos^{2}(x)=\frac{1}{3}$

$\displaystyle cos^{2}(x)-2 <=> \frac{1}{3}-2=\frac{1-6}{3}=-\frac{5}{3} =-1\frac{2}{3}$

Ответ: $\displaystyle cos^{2}(x)-2 = -1\frac{2}{3}$