What is the general solution for 1+cos(2x)=0 ?

The general solution for 1+cos(2x)=0 is x= pi/2+pin

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1+cos(2x)=0

1 + cos (2 x) = 0

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

1 + cos (2 x) = 0

Move

1

to the right side

1 + cos (2 x) = 0

Subtract

1

from both sides

1 + cos (2 x) - 1 = 0 - 1

Simplify

cos (2 x) = - 1

cos (2 x) = - 1

General solutions for

cos (2 x) = - 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 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

cos^{2} (x) - sin^{2} (x) = \frac{3}{2}

2 sin^{2} (x) + 1 = 4 sin (x)

\frac{1}{4} = cos (x)

cos (2 x) + 3 cos (x) = - 2

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

What is the general solution for 1+cos(2x)=0 ?
The general solution for 1+cos(2x)=0 is x= pi/2+pin