Question

PLS HELP, HAVE A GOOD DAY

Answer 1

Answer:

$\frac{\pi }{4} +\frac{\pi }{2} n,$  where n∈Z.

Step-by-step explanation:

2sin²x-1; ⇒ sin²x=1/2;

$\left[\begin{array}{ccc}sinx=-\frac{1}{\sqrt{2}} \\sinx=\frac{1}{\sqrt{2}} \end{array} \ = > \ \left[\begin{array}{ccc}x=(-1)^{l+1}\frac{\pi }{4}+\pi l \\x=(-1)^k\frac{\pi }{4}+\pi k \end{array} \ = > x=\frac{\pi }{4} +\frac{\pi }{2}n, \ n-Z.$

Answer 2

• 2sin²x-1=0
• 2sin²x=1
• sin²x=1/2
• sinx=1/√2

So

x:-

• nπ/2+π/4