а) ( 2 5 ) cos x + ( ( 2 5 ) − 1 ) cos x = 2 ( 2 5 ) cos x + ( 2 5 ) − cos x = 2 \left(\frac{2}{5}\right)^{\cos x}+\left(\left(\frac{2}{5}\right)^{-1}\right)^{\cos x}=2\\ \left(\frac{2}{5}\right)^{\cos x}+\left(\frac{2}{5}\right)^{-\cos x}=2 ( 5 2 ) c o s x + ( ( 5 2 ) − 1 ) c o s x = 2 ( 5 2 ) c o s x + ( 5 2 ) − c o s x = 2 ( 2 5 ) cos x = t , t > 0 \left(\frac{2}{5}\right)^{\cos x}=t,t>0 ( 5 2 ) c o s x = t , t > 0 t + 1 t = 2 ⇔ t 2 + 1 − 2 t t = 0 ⇔ { t = 1 t ≠ 0 ( 2 5 ) cos x = 1 ⇔ ( 2 5 ) cos x = ( 2 5 ) 0 cos x = 0 ⇔ x = π 2 + π k , k ∈ Z t+\frac{1}{t}=2\Leftrightarrow\frac{t^2+1-2t}{t}=0\Leftrightarrow\displaystyle \begin{cases} t=1\\ t\neq 0 \end{cases}\\ \left(\frac{2}{5}\right)^{\cos x}=1\Leftrightarrow\left(\frac{2}{5}\right)^{\cos x}=\left(\frac{2}{5}\right)^0\\ \cos x=0\Leftrightarrow x=\frac{\pi}{2}+\pi k,k\in Z t + t 1 = 2 ⇔ t t 2 + 1 − 2 t = 0 ⇔ { t = 1 t = 0 ( 5 2 ) c o s x = 1 ⇔ ( 5 2 ) c o s x = ( 5 2 ) 0 cos x = 0 ⇔ x = 2 π + π k , k ∈ Z [ − 3 π ; − 3 π 2 ] \left[-3\pi;-\frac{3\pi}{2}\right] [ − 3 π ; − 2 3 π ]
− 3 π 2 − 3 π 2 − π = − 5 π 2 -\frac{3\pi}{2}\\ -\frac{3\pi}{2}-\pi=-\frac{5\pi}{2} − 2 3 π − 2 3 π − π = − 2 5 π π 2 + π k : k ∈ Z \frac{\pi}{2}+\pi k:k\in Z 2 π + π k : k ∈ Z − 5 π 2 ; − 3 π 2 -\frac{5\pi}{2};-\frac{3\pi}{2} − 2 5 π ; − 2 3 π
