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

The general solution for 2cos(2θ)-1=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6

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2cos(2θ)-1=0,0<= θ<= 2pi

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Solution steps

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

2 cos (2 θ) = 1

Divide both sides by

2

cos (2 θ) = \frac{1}{2}

General solutions for

cos (2 θ) = \frac{1}{2}

2 θ = \frac{π}{3} + 2 πn, 2 θ = \frac{5 π}{3} + 2 πn

Solve

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

Solve

2 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{6} + πn

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

6 cos (x) + 6 sin (x) tan (x) = 12, 0 \leq x \leq 2 π

6 ∣ tan (x) ∣ = 6

cos^{2} (x) + 3 sin^{2} (x) = 1

7 sin (2 x) = - 2

sin (θ) = - \frac{( 2 )}{( 5 )}, tan (θ) > 0

What is the general solution for 2cos(2θ)-1=0,0<= θ<= 2pi ?
The general solution for 2cos(2θ)-1=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6