What is the general solution for 0=-sin(2x)-3cos(x) ?

The general solution for 0=-sin(2x)-3cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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0=-sin(2x)-3cos(x)

0 = - sin (2 x) - 3 cos (x)

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = - sin (2 x) - 3 cos (x)

Switch sides

- sin (2 x) - 3 cos (x) = 0

Rewrite using trig identities

- sin (2 x) - 3 cos (x)

Use the Double Angle identity:

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

= - 2 sin (x) cos (x) - 3 cos (x)

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

Factor

- 3 cos (x) - 2 cos (x) sin (x) : - cos (x) (2 sin (x) + 3)

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

Factor out common term

- cos (x)

= - cos (x) (3 + 2 sin (x))

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

Solving each part separately

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 sin (x) + 3 = 0 :

No Solution

2 sin (x) + 3 = 0

Move

3

to the right side

2 sin (x) + 3 = 0

Subtract

3

from both sides

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

Simplify

2 sin (x) = - 3

2 sin (x) = - 3

Divide both sides by

2

2 sin (x) = - 3

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{- 3}{2}

Simplify

sin (x) = - \frac{3}{2}

sin (x) = - \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x, sin (x) = - cos (x)

sin (x + \frac{3 π}{2}) = \frac{3}{2}

sin (θ) cos (θ) = \frac{1}{2} cos (θ)

3 sin (x) = 3 + sin (x)

cos^{2} (x) = 2 + cos (x)

What is the general solution for 0=-sin(2x)-3cos(x) ?
The general solution for 0=-sin(2x)-3cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin