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

The general solution for 2cos(θ)-sqrt(3)=0,0<= θ<= 2pi is θ= pi/6 ,θ=(11pi)/6

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2cos(θ)-sqrt(3)=0,0<= θ<= 2pi

2 cos (θ) - 3 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

Solution steps

2 cos (θ) - 3 = 0, 0 \leq θ \leq 2 π

Move

3

to the right side

2 cos (θ) = 3

Divide both sides by

2

cos (θ) = \frac{3}{2}

General solutions for

cos (θ) = \frac{3}{2}

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

3 sin (x) - 3 cos (x) = 3

tan (2 x) cos (2 x) + cot (2 x) sin (2 x) = 1

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

sin (x) = \frac{1.2}{10}

What is the general solution for 2cos(θ)-sqrt(3)=0,0<= θ<= 2pi ?
The general solution for 2cos(θ)-sqrt(3)=0,0<= θ<= 2pi is θ= pi/6 ,θ=(11pi)/6