What is the general solution for 5cos(2x)=9sin(x)+4 ?

The general solution for 5cos(2x)=9sin(x)+4 is x=(3pi)/2+2pin,x=0.10016…+2pin,x=pi-0.10016…+2pin

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5cos(2x)=9sin(x)+4

5 cos (2 x) = 9 sin (x) + 4

x = \frac{3 π}{2} + 2 πn, x = 0.10016 \dots + 2 πn, x = π - 0.10016 \dots + 2 πn

Solution steps

5 cos (2 x) = 9 sin (x) + 4

Subtract

9 sin (x) + 4

from both sides

5 cos (2 x) - 9 sin (x) - 4 = 0

Rewrite using trig identities

1 - 10 sin^{2} (x) - 9 sin (x) = 0

Solve by substitution

sin (x) = - 1, sin (x) = \frac{1}{10}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{10} : x = arcsin (\frac{1}{10}) + 2 πn, x = π - arcsin (\frac{1}{10}) + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{1}{10}) + 2 πn, x = π - arcsin (\frac{1}{10}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{2} + 2 πn, x = 0.10016 \dots + 2 πn, x = π - 0.10016 \dots + 2 πn

- 3 cos (x) - 1 = cos (2 x)

- 2 tan^{2} (θ) - 4 tan (θ) + 2 = 3 tan (θ) + 5

csc (3 x) = 1

x, cos (x + y) + cos (x) = 0

cos (\frac{π t}{2}) \frac{π}{2} = 0

What is the general solution for 5cos(2x)=9sin(x)+4 ?
The general solution for 5cos(2x)=9sin(x)+4 is x=(3pi)/2+2pin,x=0.10016…+2pin,x=pi-0.10016…+2pin