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

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

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4cos^2(x)-1=0,0<= x<2pi

4 cos^{2} (x) - 1 = 0, 0 \leq x < 2 π

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

Solution steps

4 cos^{2} (x) - 1 = 0, 0 \leq x < 2 π

Solve by substitution

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

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

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

Combine all the solutions

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

y, e^{2 x} = sin (x + 3 y)

7 sin (2 x) = 7 cos (x)

cos (θ) = - 0.17

3 sin (2 x) - 2 sin (x) = 0

cos^{2} (θ) = 1

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