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

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

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cos(3x)=-1,0<= x<= 2pi

cos (3 x) = - 1, 0 \leq x \leq 2 π

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

Degrees

x = 6 0^{\circ}, x = 18 0^{\circ}, x = 30 0^{\circ}

Solution steps

cos (3 x) = - 1, 0 \leq x \leq 2 π

General solutions for

cos (3 x) = - 1

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}

3 x = π + 2 πn

3 x = π + 2 πn

Solve

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

3 x = π + 2 πn

Divide both sides by

3

3 x = π + 2 πn

Divide both sides by

3

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

Simplify

x = \frac{π}{3} + \frac{2 πn}{3}

x = \frac{π}{3} + \frac{2 πn}{3}

x = \frac{π}{3} + \frac{2 πn}{3}

Solutions for the range

0 \leq x \leq 2 π

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

sin (x) = - 3 cos (x)

4 sin^{2} (x) + 2 cos^{2} (x) = 3, 0 \leq x \leq 2 π

sin (x) = - 5

sin (x) = 10

tan (A) = - \frac{3}{91}

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