Answer:
[tex]\frac{\pi }{4} +\frac{\pi }{2} n,[/tex] where n∈Z.
Step-by-step explanation:
2sin²x-1; ⇒ sin²x=1/2;
[tex]\left[\begin{array}{ccc}sinx=-\frac{1}{\sqrt{2}} \\sinx=\frac{1}{\sqrt{2}} \end{array} \ = > \ \left[\begin{array}{ccc}x=(-1)^{l+1}\frac{\pi }{4}+\pi l \\x=(-1)^k\frac{\pi }{4}+\pi k \end{array} \ = > x=\frac{\pi }{4} +\frac{\pi }{2}n, \ n-Z.[/tex]
