What is the general solution for 4cos^2(θ)=1,0<= θ<2pi ?

The general solution for 4cos^2(θ)=1,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3 ,θ=(2pi)/3 ,θ=(4pi)/3

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4cos^2(θ)=1,0<= θ<2pi

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Solution steps

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{5 π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

cos (\frac{x}{2} + \frac{π}{3}) = \frac{2}{2}

tan (θ) = 8.2

2 sin (x) = 4 sin (x) + 1

- 1 - cos (θ) = - 3 cos (θ)

3 \cdot cos (θ) = 2 - sin (θ)

What is the general solution for 4cos^2(θ)=1,0<= θ<2pi ?
The general solution for 4cos^2(θ)=1,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3 ,θ=(2pi)/3 ,θ=(4pi)/3