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

Q: What is the general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi ?

The general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2

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2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi

Solution

2 sin^{2} (θ) - 2 = cos^{2} (θ), 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Degrees

θ = 9 0^{\circ}, θ = 27 0^{\circ}

Solution steps

2 sin^{2} (θ) - 2 = cos^{2} (θ), 0 \leq θ \leq 2 π

Subtract

cos^{2} (θ)

from both sides

2 sin^{2} (θ) - 2 - cos^{2} (θ) = 0

Rewrite using trig identities

- 2 - cos^{2} (θ) + 2 sin^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

- 2 - (1 - sin^{2} (θ)) + 2 sin^{2} (θ) : 3 sin^{2} (θ) - 3

- 2 - (1 - sin^{2} (θ)) + 2 sin^{2} (θ)

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

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

Simplify

- 2 - 1 + sin^{2} (θ) + 2 sin^{2} (θ) : 3 sin^{2} (θ) - 3

- 2 - 1 + sin^{2} (θ) + 2 sin^{2} (θ)

Add similar elements:

sin^{2} (θ) + 2 sin^{2} (θ) = 3 sin^{2} (θ)

= - 2 - 1 + 3 sin^{2} (θ)

Subtract the numbers:

- 2 - 1 = - 3

= 3 sin^{2} (θ) - 3

= 3 sin^{2} (θ) - 3

= 3 sin^{2} (θ) - 3

- 3 + 3 sin^{2} (θ) = 0

Solve by substitution

- 3 + 3 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 3 + 3 u^{2} = 0

- 3 + 3 u^{2} = 0 : u = 1, u = - 1

- 3 + 3 u^{2} = 0

Move

3

to the right side

- 3 + 3 u^{2} = 0

Add

3

to both sides

- 3 + 3 u^{2} + 3 = 0 + 3

Simplify

3 u^{2} = 3

3 u^{2} = 3

Divide both sides by

3

3 u^{2} = 3

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{3}{3}

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 = sin (θ)

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1, 0 \leq θ \leq 2 π : θ = \frac{π}{2}

sin (θ) = 1, 0 \leq θ \leq 2 π

General solutions for

sin (θ) = 1

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 = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{2}

sin (θ) = - 1, 0 \leq θ \leq 2 π : θ = \frac{3 π}{2}

sin (θ) = - 1, 0 \leq θ \leq 2 π

General solutions for

sin (θ) = - 1

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 = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{3 π}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi ?
The general solution for 2sin^2(θ)-2=cos^2(θ),0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2