Asked by naledi
general solution of sin2x = 4cos2x
Answers
Answered by
oobleck
recall that sinθ = cosθ at θ = π/4 + kπ
since θ = 2x, we have
x = π/8 + kπ/2
since θ = 2x, we have
x = π/8 + kπ/2
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