What is the general solution for sin^2(θ)-1=0,0<= θ<= 2pi ?

Q: What is the general solution for sin^2(θ)-1=0,0<= θ<= 2pi ?

The general solution for sin^2(θ)-1=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2

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sin^2(θ)-1=0,0<= θ<= 2pi

Solution

sin^{2} (θ) - 1 = 0, 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

sin^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

Solve by substitution

sin^{2} (θ) - 1 = 0

Let:

sin (θ) = u

u^{2} - 1 = 0

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

u^{2} - 1 = 0

Move

1

to the right side

u^{2} - 1 = 0

Add

1

to both sides

u^{2} - 1 + 1 = 0 + 1

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}

Popular Examples

cos(3θ)=(sqrt(3))/2

cos (3 θ) = \frac{3}{2}

tan^2(θ)-1=0,pi<= θ<= 2pi

tan^{2} (θ) - 1 = 0, π \leq θ \leq 2 π

sin(y)=-1/2

sin (y) = - \frac{1}{2}

cos^{22}(x)-2sin^2(x)-1=0

cos^{22} (x) - 2 sin^{2} (x) - 1 = 0

tan(θ)=(3sqrt(5))/2

tan (θ) = \frac{3 5}{2}

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)-1=0,0<= θ<= 2pi ?
The general solution for sin^2(θ)-1=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2