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

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

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cos(x)=0,-pi<= x<= pi

cos (x) = 0, - π \leq x \leq π

x = \frac{π}{2}, x = - \frac{π}{2}

Degrees

x = 9 0^{\circ}, x = - 9 0^{\circ}

Solution steps

cos (x) = 0, - π \leq x \leq π

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

- π \leq x \leq π

x = \frac{π}{2}, x = - \frac{π}{2}

sin^{2} (x) + cos (x) = 2

sin (θ) = - \frac{4}{5}, \frac{7 π}{2} \leq θ < 4 π, cos (2 θ)

cos (2 x) - 2 sin^{2} (x) - sin (x) = 1

- 8 sin (θ) = - 3

sin (x) + 1 = 1

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