2/(sqrt(3))cos(3θ)=1,0<= θ<= 2pi

Solution

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

θ = \frac{π}{18}, θ = \frac{11 π}{18}, θ = \frac{13 π}{18}, θ = \frac{23 π}{18}, θ = \frac{25 π}{18}, θ = \frac{35 π}{18}

Solution steps

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

Simplify

\frac{2}{3} : \frac{2 3}{3}

\frac{2 3}{3} cos (3 θ) = 1

Multiply both sides by

3

23 cos (3 θ) = 3

Divide both sides by

23

cos (3 θ) = \frac{3}{2}

General solutions for

cos (3 θ) = \frac{3}{2}

3 θ = \frac{π}{6} + 2 πn, 3 θ = \frac{11 π}{6} + 2 πn

Solve

3 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{11 π}{18} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{18}, θ = \frac{11 π}{18}, θ = \frac{13 π}{18}, θ = \frac{23 π}{18}, θ = \frac{25 π}{18}, θ = \frac{35 π}{18}

8 cos (B) + 52 = 0

sec (\frac{x}{2}) = 2

sin (5 x) = \frac{3}{2}

sin^{2} (θ) - cos^{2} (θ) = - 1

cos^{2} (x) - 4 cos (x) = 0