What is the general solution for 2sin(2x)+sqrt(2)=0 ?

The general solution for 2sin(2x)+sqrt(2)=0 is x=(5pi)/8+pin,x=(7pi)/8+pin

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2sin(2x)+sqrt(2)=0

Solution

2 sin (2 x) + 2 = 0

Solution

x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

Degrees

x = 112. 5^{\circ} + 18 0^{\circ} n, x = 157. 5^{\circ} + 18 0^{\circ} n

Solution steps

2 sin (2 x) + 2 = 0

Move

2

to the right side

2 sin (2 x) + 2 = 0

Subtract

2

from both sides

2 sin (2 x) + 2 - 2 = 0 - 2

Simplify

2 sin (2 x) = - 2

2 sin (2 x) = - 2

Divide both sides by

2

2 sin (2 x) = - 2

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} = \frac{- 2}{2}

Simplify

sin (2 x) = - \frac{2}{2}

sin (2 x) = - \frac{2}{2}

General solutions for

sin (2 x) = - \frac{2}{2}

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 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

2 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

Solve

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

2 x = \frac{5 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

Apply the fraction rule:

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

= \frac{5 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

Solve

2 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{8} + πn

2 x = \frac{7 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{7 π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2} : \frac{7 π}{8} + πn

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{4}}{2} = \frac{7 π}{8}

\frac{\frac{7 π}{4}}{2}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{7 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

Popular Examples

2cos^2(θ)-3sin(θ)+2=0

2 cos^{2} (θ) - 3 sin (θ) + 2 = 0

sin(x)+sin(2x)+sin(3x)=0

sin (x) + sin (2 x) + sin (3 x) = 0

sin^2(x)-cos^2(x)=1+cos(x)

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

36cos^2(x)-9=0

36 cos^{2} (x) - 9 = 0

sin(x)= 5/6

sin (x) = \frac{5}{6}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2x)+sqrt(2)=0 ?
The general solution for 2sin(2x)+sqrt(2)=0 is x=(5pi)/8+pin,x=(7pi)/8+pin