What is the general solution for cos(2θ)-cos^2(θ)=0 ?

The general solution for cos(2θ)-cos^2(θ)=0 is θ=2pin,θ=pi+2pin

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cos(2θ)-cos^2(θ)=0

cos (2 θ) - cos^{2} (θ) = 0

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 θ) - cos^{2} (θ) = 0

Rewrite using trig identities

cos (2 θ) - cos^{2} (θ)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

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

= cos (2 θ) - (cos (2 θ) + sin^{2} (θ))

Simplify

cos (2 θ) - (cos (2 θ) + sin^{2} (θ)) : - sin^{2} (θ)

cos (2 θ) - (cos (2 θ) + sin^{2} (θ))

- (cos (2 θ) + sin^{2} (θ)) : - cos (2 θ) - sin^{2} (θ)

- (cos (2 θ) + sin^{2} (θ))

Distribute parentheses

= - (cos (2 θ)) - (sin^{2} (θ))

Apply minus-plus rules

+ (- a) = - a

= - cos (2 θ) - sin^{2} (θ)

= cos (2 θ) - cos (2 θ) - sin^{2} (θ)

Add similar elements:

cos (2 θ) - cos (2 θ) = 0

= - sin^{2} (θ)

= - sin^{2} (θ)

- sin^{2} (θ) = 0

Divide both sides by

- 1

- sin^{2} (θ) = 0

Divide both sides by

- 1

- sin^{2} (θ) = 0

Divide both sides by

- 1

\frac{- sin ^{2} ( θ )}{- 1} = \frac{0}{- 1}

Simplify

sin^{2} (θ) = 0

sin^{2} (θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

3 tan^{2} (x) + 5 tan (x) - 4 = 0

cos (2 x) \cdot 2 = 0

csc^{3} (x) + csc^{2} (x) = 4 csc (x) + 4

2 sin^{2} (x) - 5 sin (x) - 3 = 0

6 + 6 sin (θ) = 4 cos^{2} (θ)

What is the general solution for cos(2θ)-cos^2(θ)=0 ?
The general solution for cos(2θ)-cos^2(θ)=0 is θ=2pin,θ=pi+2pin