What is the general solution for 2cos(θ)-3=-5,0<= θ<2pi ?

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

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2cos(θ)-3=-5,0<= θ<2pi

2 cos (θ) - 3 = - 5, 0 \leq θ < 2 π

θ = π

Degrees

θ = 18 0^{\circ}

Solution steps

2 cos (θ) - 3 = - 5, 0 \leq θ < 2 π

Move

3

to the right side

2 cos (θ) - 3 = - 5

Add

3

to both sides

2 cos (θ) - 3 + 3 = - 5 + 3

Simplify

2 cos (θ) = - 2

2 cos (θ) = - 2

Divide both sides by

2

2 cos (θ) = - 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 2}{2}

Simplify

cos (θ) = - 1

cos (θ) = - 1

General solutions for

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = π

2 tan (x) + 1 = 0

2 cos (3 x) + cos (2 x) + 1 = 0, 0 \leq x \leq 2 π

tan^{2} (x) = 3

tan (x) csc^{2} (x) - 2 tan (x) = 0

5 sin (2 x) = cos (x)

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