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

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

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

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

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

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

Factor

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

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

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from 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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

6 cos^{2} (x) + 5 cos (x) + 1 = 0

tan^{4} (x) + 1 = 4 tan^{2} (x) - 2

tan^{2} (x) = sec (x)

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

sin (θ) = \frac{4}{5}, cos (θ)

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