Find the solution of the equation log (cos2x) [Base sinx] = 2 in the interval [-PI/2,PI/2].

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Find the solutions to a trig equation between 0 and 2pi

Let A= {x∈(0,π)-{π/2} ∶ 〖log〗_(2/π) |sinx|+〖log〗_(2/π) |cosx|=2} and B = {x≥0∶ √x (√x- 4)- 3|√x-2|+

[IIT 1980] Find the solution of the given trigonometric equation in the interval [0,2*PI].

Solve the Equation sin⁡(2x-π/2)=√2/2 on the Interval [0,2π).

Solve the Trig equation sin x + cos x = -sqrt(2) on the interval [0, 2pi)

Find the smallest positive x satisfying log (sinx) [Base cosx] + log (cosx) [Base sinx] = 2.

Evaluate integrate 0 to pi /2 [log(sin x) - log(2cos x)] dx from 0 to pi/2 | ∫ log(sinx) -log(2cosx)

Solving TRIGONOMETRY Equation sin(2x) = sqrt(3)cos(x) In Interval [0, 2pi)