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

The general solution for -sin(2x)=-2sin(x) is x=2pin,x=pi+2pin

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Popular >

-sin(2x)=-2sin(x)

Solution

- sin (2 x) = - 2 sin (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

- sin (2 x) = - 2 sin (x)

Subtract

- 2 sin (x)

from both sides

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

Rewrite using trig identities

- sin (2 x) + 2 sin (x)

Use the Double Angle identity:

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

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

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

Factor

2 sin (x) - 2 cos (x) sin (x) : - 2 sin (x) (cos (x) - 1)

2 sin (x) - 2 cos (x) sin (x)

Factor out common term

- 2 sin (x)

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

- 2 sin (x) (cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or cos (x) - 1 = 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

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

cos (x) - 1 + 1 = 0 + 1

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 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

Combine all the solutions

x = 2 πn, x = π + 2 πn

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sin (8 x + 2) = cos (6 x - 10)

cos(2θ-pi/2)=-1,0<= θ<= 2pi

cos (2 θ - \frac{π}{2}) = - 1, 0 \leq θ \leq 2 π

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

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

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

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

(sqrt(3))/2 =sin(arcsin(0)+c_{1})

\frac{3}{2} = sin (arcsin (0) + c_{1})

Frequently Asked Questions (FAQ)

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