What is the general solution for 0.4=0.4cos^2(θ) ?

The general solution for 0.4=0.4cos^2(θ) is θ=2pin,θ=pi+2pin

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0.4=0.4cos^2(θ)

0.4 = 0.4 cos^{2} (θ)

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

0.4 = 0.4 cos^{2} (θ)

Switch sides

0.4 cos^{2} (θ) = 0.4

Solve by substitution

0.4 cos^{2} (θ) = 0.4

Let:

cos (θ) = u

0.4 u^{2} = 0.4

0.4 u^{2} = 0.4 : u = 1, u = - 1

0.4 u^{2} = 0.4

Divide both sides by

0.4

0.4 u^{2} = 0.4

Divide both sides by

0.4

\frac{0.4 u ^{2}}{0.4} = \frac{0.4}{0.4}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = - 1

cos (θ) = 1, cos (θ) = - 1

cos (θ) = 1 : θ = 2 πn

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 = 0 + 2 πn

θ = 0 + 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

cos (θ) = - 1 : θ = π + 2 πn

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

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

sin (α) = \frac{15}{17}

4 sin^{2} (x) cot (x) - cot (x) = 0, 0 \leq x \leq 2 π

14 - sin (θ) = cos (2 θ)

csc (θ) = - \frac{5}{13}

cos (z) = 3

What is the general solution for 0.4=0.4cos^2(θ) ?
The general solution for 0.4=0.4cos^2(θ) is θ=2pin,θ=pi+2pin