What is the general solution for 2cos(x)sin(x)+9sin(x)=0 ?

The general solution for 2cos(x)sin(x)+9sin(x)=0 is x=2pin,x=pi+2pin

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2cos(x)sin(x)+9sin(x)=0

2 cos (x) sin (x) + 9 sin (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 cos (x) sin (x) + 9 sin (x) = 0

Factor

2 cos (x) sin (x) + 9 sin (x) : sin (x) (2 cos (x) + 9)

2 cos (x) sin (x) + 9 sin (x)

Factor out common term

sin (x)

= sin (x) (2 cos (x) + 9)

sin (x) (2 cos (x) + 9) = 0

Solving each part separately

sin (x) = 0 or 2 cos (x) + 9 = 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

2 cos (x) + 9 = 0 :

No Solution

2 cos (x) + 9 = 0

Move

9

to the right side

2 cos (x) + 9 = 0

Subtract

9

from both sides

2 cos (x) + 9 - 9 = 0 - 9

Simplify

2 cos (x) = - 9

2 cos (x) = - 9

Divide both sides by

2

2 cos (x) = - 9

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{- 9}{2}

Simplify

cos (x) = - \frac{9}{2}

cos (x) = - \frac{9}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

1 - sin (x) = 0

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

4 sec^{2} (x) + 2 tan^{2} (x) - 6 = 0

tan (θ) = - \frac{1}{2}

4 sin^{2} (θ) - 2 = 0

What is the general solution for 2cos(x)sin(x)+9sin(x)=0 ?
The general solution for 2cos(x)sin(x)+9sin(x)=0 is x=2pin,x=pi+2pin