What is the general solution for (2cos(2θ))=2 ?

The general solution for (2cos(2θ))=2 is θ=pin

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(2cos(2θ))=2

(2 cos (2 θ)) = 2

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

(2 cos (2 θ)) = 2

Divide both sides by

2

2 cos (2 θ) = 2

Divide both sides by

2

\frac{2 cos ( 2 θ )}{2} = \frac{2}{2}

Simplify

cos (2 θ) = 1

cos (2 θ) = 1

General solutions for

cos (2 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}

2 θ = 0 + 2 πn

2 θ = 0 + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

Divide both sides by

2

2 θ = 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2}

Simplify

θ = πn

θ = πn

θ = πn

5 sin^{2} (x) = 9 tan (x)

cos (θ) = \frac{5}{4}

cos (x) = sin (2 x), 0 \leq x \leq 2 π

b, g = sin (b cos (d e tan (c)))

tan (θ) (sin (θ) - 1) = 0

What is the general solution for (2cos(2θ))=2 ?
The general solution for (2cos(2θ))=2 is θ=pin