What is the general solution for sin^2(θ)=(cos^2(θ)-1)/2 ?

The general solution for sin^2(θ)=(cos^2(θ)-1)/2 is θ=2pin,θ=pi+2pin

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sin^2(θ)=(cos^2(θ)-1)/2

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

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

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

Subtract

\frac{cos ^{2} ( θ ) - 1}{2}

from both sides

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

Simplify

sin^{2} (θ) - \frac{cos ^{2} ( θ ) - 1}{2} : \frac{2 sin ^{2} ( θ ) - cos ^{2} ( θ ) + 1}{2}

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

Convert element to fraction:

sin^{2} (θ) = \frac{sin ^{2} ( θ ) 2}{2}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Expand

sin^{2} (θ) \cdot 2 - (cos^{2} (θ) - 1) : sin^{2} (θ) \cdot 2 - cos^{2} (θ) + 1

sin^{2} (θ) \cdot 2 - (cos^{2} (θ) - 1)

= 2 sin^{2} (θ) - (cos^{2} (θ) - 1)

- (cos^{2} (θ) - 1) : - cos^{2} (θ) + 1

- (cos^{2} (θ) - 1)

Distribute parentheses

= - (cos^{2} (θ)) - (- 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos^{2} (θ) + 1

= sin^{2} (θ) \cdot 2 - cos^{2} (θ) + 1

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin^{2} (θ) - cos^{2} (θ) + 1 = 0

Rewrite using trig identities

1 - cos^{2} (θ) + 2 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 2 sin^{2} (θ) + sin^{2} (θ)

Simplify

= 3 sin^{2} (θ)

3 sin^{2} (θ) = 0

Divide both sides by

3

3 sin^{2} (θ) = 0

Divide both sides by

3

3 sin^{2} (θ) = 0

Divide both sides by

3

\frac{3 sin ^{2} ( θ )}{3} = \frac{0}{3}

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

tan (x) = \frac{3}{10}

y = 1 + sin (\frac{π x}{4}), x

sec (\frac{x}{2}) = 2

arccos (2 x) + arccos (x) = \frac{π}{3}

tan (2 x) = \frac{500}{350 - 75}

What is the general solution for sin^2(θ)=(cos^2(θ)-1)/2 ?
The general solution for sin^2(θ)=(cos^2(θ)-1)/2 is θ=2pin,θ=pi+2pin