What is the general solution for sin(2θ)+1=0 ?

The general solution for sin(2θ)+1=0 is θ=(3pi)/4+pin

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sin(2θ)+1=0

sin (2 θ) + 1 = 0

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 θ) + 1 = 0

Move

1

to the right side

sin (2 θ) + 1 = 0

Subtract

1

from both sides

sin (2 θ) + 1 - 1 = 0 - 1

Simplify

sin (2 θ) = - 1

sin (2 θ) = - 1

General solutions for

sin (2 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}

2 θ = \frac{3 π}{2} + 2 πn

2 θ = \frac{3 π}{2} + 2 πn

Solve

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

x, cos (x) = cos (y)

3 tanh^{2} (θ) = 5 sech (θ) + 1

6 cos (x) + 6 sin (x) tan (x) = 12

cos (θ) = \frac{8}{17}, 27 0^{\circ} < θ < 36 0^{\circ}

2 csc (θ) - 3 = 0

What is the general solution for sin(2θ)+1=0 ?
The general solution for sin(2θ)+1=0 is θ=(3pi)/4+pin