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

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

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sqrt(2sin(2x))=0

Solution

2 sin (2 x) = 0

Solution

x = πn, x = \frac{π}{2} + πn

Radians

x = 0 + πn, x = \frac{π}{2} + πn

Solution steps

2 sin (2 x) = 0

Square both sides:

2 sin (2 x) = 0

2 sin (2 x) = 0

(2 sin (2 x))^{2} = 0^{2}

Expand

(2 sin (2 x))^{2} : 2 sin (2 x)

(2 sin (2 x))^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((2 sin (2 x))^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (2 sin (2 x))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2 sin (2 x)

Expand

0^{2} : 0

0^{2}

Apply rule

0^{a} = 0

= 0

2 sin (2 x) = 0

2 sin (2 x) = 0

Solve

2 sin (2 x) = 0 : sin (2 x) = 0

2 sin (2 x) = 0

Divide both sides by

2

2 sin (2 x) = 0

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} = \frac{0}{2}

Simplify

sin (2 x) = 0

sin (2 x) = 0

sin (2 x) = 0

Verify Solutions:

sin (2 x) = 0

True

Check the solutions by plugging them into

2 sin (2 x) = 0

Remove the ones that don't agree with the equation.

Plug in

sin (2 x) = 0 :

True

2 \cdot 0 = 0

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

Apply rule

0 = 0

= 0

0 = 0

True

The solution is

sin (2 x) = 0

General solutions for

sin (2 x) = 0

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 = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Popular Examples

2cos(x)+cos(2x)=0

2 cos (x) + cos (2 x) = 0

solvefor x,-sin(x+y)=0 solve for

x, - sin (x + y) = 0

cot(θ)=-1/2

cot (θ) = - \frac{1}{2}

sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

sin^2(t)=cos^2(t)

sin^{2} (t) = cos^{2} (t)

Frequently Asked Questions (FAQ)

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