What is the general solution for 5sin(2x)+9sin(x)=0 ?

The general solution for 5sin(2x)+9sin(x)=0 is x=2pin,x=pi+2pin,x=2.69056…+2pin,x=-2.69056…+2pin

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Popular >

5sin(2x)+9sin(x)=0

Solution

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

Solution

x = 2 πn, x = π + 2 πn, x = 2.69056 \dots + 2 πn, x = - 2.69056 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 154.15806 \dots^{\circ} + 36 0^{\circ} n, x = - 154.15806 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

5 sin (2 x) + 9 sin (x)

Use the Double Angle identity:

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

= 5 \cdot 2 sin (x) cos (x) + 9 sin (x)

Simplify

= 10 sin (x) cos (x) + 9 sin (x)

9 sin (x) + 10 cos (x) sin (x) = 0

Factor

9 sin (x) + 10 cos (x) sin (x) : sin (x) (10 cos (x) + 9)

9 sin (x) + 10 cos (x) sin (x)

Factor out common term

sin (x)

= sin (x) (9 + 10 cos (x))

sin (x) (10 cos (x) + 9) = 0

Solving each part separately

sin (x) = 0 or 10 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

10 cos (x) + 9 = 0 : x = arccos (- \frac{9}{10}) + 2 πn, x = - arccos (- \frac{9}{10}) + 2 πn

10 cos (x) + 9 = 0

Move

9

to the right side

10 cos (x) + 9 = 0

Subtract

9

from both sides

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

Simplify

10 cos (x) = - 9

10 cos (x) = - 9

Divide both sides by

10

10 cos (x) = - 9

Divide both sides by

10

\frac{10 cos ( x )}{10} = \frac{- 9}{10}

Simplify

cos (x) = - \frac{9}{10}

cos (x) = - \frac{9}{10}

Apply trig inverse properties

cos (x) = - \frac{9}{10}

General solutions for

cos (x) = - \frac{9}{10}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{9}{10}) + 2 πn, x = - arccos (- \frac{9}{10}) + 2 πn

x = arccos (- \frac{9}{10}) + 2 πn, x = - arccos (- \frac{9}{10}) + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arccos (- \frac{9}{10}) + 2 πn, x = - arccos (- \frac{9}{10}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 2.69056 \dots + 2 πn, x = - 2.69056 \dots + 2 πn

Popular Examples

4sin(x)-6cos(x)=3

4 sin (x) - 6 cos (x) = 3

4cos(x)=cos^3(x)

4 cos (x) = cos^{3} (x)

cos^2(a)= 3/(5sin(a))

cos^{2} (a) = \frac{3}{5 sin ( a )}

3cos(x)=2sec(x)-5

3 cos (x) = 2 sec (x) - 5

sin(a/2)=(1/2)sin(a)

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

Frequently Asked Questions (FAQ)

What is the general solution for 5sin(2x)+9sin(x)=0 ?
The general solution for 5sin(2x)+9sin(x)=0 is x=2pin,x=pi+2pin,x=2.69056…+2pin,x=-2.69056…+2pin