What is the general solution for 4cos(2θ)=4sin(2θ) ?

The general solution for 4cos(2θ)=4sin(2θ) is θ= pi/8+(pin)/2

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4cos(2θ)=4sin(2θ)

4 cos (2 θ) = 4 sin (2 θ)

θ = \frac{π}{8} + \frac{πn}{2}

Solution steps

4 cos (2 θ) = 4 sin (2 θ)

Rewrite using trig identities

4 tan (2 θ) = 4

Divide both sides by

4

tan (2 θ) = 1

General solutions for

tan (2 θ) = 1

2 θ = \frac{π}{4} + πn

Solve

2 θ = \frac{π}{4} + πn : θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

tan (x) = - 3, 0 \leq x \leq 2 π

5 tan^{2} (A) - 7 = 8

tan (x) = \frac{28}{16}

2 sin^{2} (x) = 3 - sin (x)

9 tan (2 x) - 18 cos (x) = 0

What is the general solution for 4cos(2θ)=4sin(2θ) ?
The general solution for 4cos(2θ)=4sin(2θ) is θ= pi/8+(pin)/2