Question

2 (sin square x - cos square x)=0​

Answer 1

given

=> 2(sin²x - cos²x) = 0

=> (sinx + cosx) (sinx - cosx) = 0

=> sinx + cosx = 0 or sinx - cosx = 0

=> cotx = -1 or cotx = 1

=> x = -π/4 or x = π/4

for general solution

=> x = 2nπ - π/4 or x = 2nπ + π/4

where n = ±1,±2,±3,..........

Hope it may Help you.

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