What is the general solution for -8cos(2x)=0,(-pi,pi) ?

The general solution for -8cos(2x)=0,(-pi,pi) is x= pi/4 ,x=(3pi)/4 ,x=-(3pi)/4 ,x=-pi/4

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-8cos(2x)=0,(-pi,pi)

- 8 cos (2 x) = 0, (- π, π)

x = \frac{π}{4}, x = \frac{3 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}

Solution steps

- 8 cos (2 x) = 0, (- π, π)

Divide both sides by

- 8

cos (2 x) = 0

General solutions for

cos (2 x) = 0

2 x = \frac{π}{2} + 2 πn, 2 x = \frac{3 π}{2} + 2 πn

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Solutions for the range

(- π, π)

x = \frac{π}{4}, x = \frac{3 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}

1.66 = sin (x)

sin (x) = \frac{3 π}{2}

2 sec^{2} (x) = 3 - tan (x), 0 \leq x \leq 2 π

1 + cos (2 θ) = - 3 cos (θ)

tan (θ) = \frac{6}{0}

What is the general solution for -8cos(2x)=0,(-pi,pi) ?
The general solution for -8cos(2x)=0,(-pi,pi) is x= pi/4 ,x=(3pi)/4 ,x=-(3pi)/4 ,x=-pi/4