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

The general solution for 2sin(2x)=sin^2(x) is x=2pin,x=pi+2pin,x=1.32581…+pin

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Popular >

2sin(2x)=sin^2(x)

Solution

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

Solution

x = 2 πn, x = π + 2 πn, x = 1.32581 \dots + πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 75.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Simplify

= - sin^{2} (x) + 4 sin (x) cos (x)

- sin^{2} (x) + 4 cos (x) sin (x) = 0

Factor

- sin^{2} (x) + 4 cos (x) sin (x) : sin (x) (- sin (x) + 4 cos (x))

- sin^{2} (x) + 4 cos (x) sin (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

= - sin (x) sin (x) + 4 sin (x) cos (x)

Factor out common term

sin (x)

= sin (x) (- sin (x) + 4 cos (x))

sin (x) (- sin (x) + 4 cos (x)) = 0

Solving each part separately

sin (x) = 0 or - sin (x) + 4 cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

- sin (x) + 4 cos (x) = 0 : x = arctan (4) + πn

- sin (x) + 4 cos (x) = 0

Rewrite using trig identities

- sin (x) + 4 cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{- sin ( x ) + 4 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

- \frac{sin ( x )}{cos ( x )} + 4 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

- tan (x) + 4 = 0

- tan (x) + 4 = 0

Move

4

to the right side

- tan (x) + 4 = 0

Subtract

4

from both sides

- tan (x) + 4 - 4 = 0 - 4

Simplify

- tan (x) = - 4

- tan (x) = - 4

Divide both sides by

- 1

- tan (x) = - 4

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 4}{- 1}

Simplify

tan (x) = 4

tan (x) = 4

Apply trig inverse properties

tan (x) = 4

General solutions for

tan (x) = 4

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (4) + πn

x = arctan (4) + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arctan (4) + πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 1.32581 \dots + πn

Popular Examples

5sin^2(x)=3sin(x)+2

5 sin^{2} (x) = 3 sin (x) + 2

sin^2(x)=-(1/2)

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

solvefor α,sin^2(3α)=0.56 solve for

α, sin^{2} (3 α) = 0.56

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

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

tan(pi/2-x)+2tan(pi/2+x)=1

tan (\frac{π}{2} - x) + 2 tan (\frac{π}{2} + x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2x)=sin^2(x) ?
The general solution for 2sin(2x)=sin^2(x) is x=2pin,x=pi+2pin,x=1.32581…+pin