What is the general solution for sin^2(2θ)=1,-180<= θ<= 180 ?

Q: What is the general solution for sin^2(2θ)=1,-180<= θ<= 180 ?

The general solution for sin^2(2θ)=1,-180<= θ<= 180 is θ=45,θ=-135,θ=135,θ=-45

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sin^2(2θ)=1,-180<= θ<= 180

Solution

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

Solution

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

Radians

θ = \frac{π}{4}, θ = - \frac{3 π}{4}, θ = \frac{3 π}{4}, θ = - \frac{π}{4}

Solution steps

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

Solve by substitution

sin^{2} (2 θ) = 1

Let:

sin (2 θ) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 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 (2 θ)

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

General solutions for

sin (2 θ) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 9 0^{\circ} + 36 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

Solve

2 θ = 9 0^{\circ} + 36 0^{\circ} n : θ = 4 5^{\circ} + 18 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 9 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 4 5^{\circ} + 18 0^{\circ} n

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{9 0 ^{\circ}}{2} = 4 5^{\circ}

\frac{9 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 4 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

Solutions for the range

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 13 5^{\circ}, θ = - 4 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

General solutions for

sin (2 θ) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 27 0^{\circ} + 36 0^{\circ} n

2 θ = 27 0^{\circ} + 36 0^{\circ} n

Solve

2 θ = 27 0^{\circ} + 36 0^{\circ} n : θ = 13 5^{\circ} + 18 0^{\circ} n

2 θ = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{54 0 ^{\circ}}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

Solutions for the range

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 13 5^{\circ}, θ = - 4 5^{\circ}

Combine all the solutions

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(2θ)=1,-180<= θ<= 180 ?
The general solution for sin^2(2θ)=1,-180<= θ<= 180 is θ=45,θ=-135,θ=135,θ=-45