Question: * Solve the equation on the interval [ 0, 2pi ) 1. sin 4x = (square root of 3) / 2 (how to get the

* Solve the equation on the interval [ 0, 2pi )

1. sin 4x = (square root of 3) / 2 (how to get the answer for pi/12, pi/6, 2pi/3, 7pi/12, 7pi/6, 13pi/12, 5pi/3, 19pi/12)

2. ( cot^2 x )( cos x ) = cot^2 x (how to get the answer for pi/2, 3pi/2)

3. ( sec^2 x ) - 2 = tan^2 x (how to get the answer for "no solution")

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