What is the general solution for 9cos(2x)=10sin(x)+1 ?

The general solution for 9cos(2x)=10sin(x)+1 is x=(3pi)/2+2pin,x=0.46055…+2pin,x=pi-0.46055…+2pin

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9cos(2x)=10sin(x)+1

9 cos (2 x) = 10 sin (x) + 1

x = \frac{3 π}{2} + 2 πn, x = 0.46055 \dots + 2 πn, x = π - 0.46055 \dots + 2 πn

Solution steps

9 cos (2 x) = 10 sin (x) + 1

Subtract

10 sin (x) + 1

from both sides

9 cos (2 x) - 10 sin (x) - 1 = 0

Rewrite using trig identities

8 - 10 sin (x) - 18 sin^{2} (x) = 0

Solve by substitution

sin (x) = - 1, sin (x) = \frac{4}{9}

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

sin (x) = \frac{4}{9} : x = arcsin (\frac{4}{9}) + 2 πn, x = π - arcsin (\frac{4}{9}) + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{4}{9}) + 2 πn, x = π - arcsin (\frac{4}{9}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{2} + 2 πn, x = 0.46055 \dots + 2 πn, x = π - 0.46055 \dots + 2 πn

tan^{2} (x) - 1 = 0, 0 \leq x \leq \frac{π}{2}

10 cos (B) + 6 = 4 cos (B) + 4

2 sin (x) + 2 sin (x) tan (x) = 0

y, x = x (1 + tan (y))

- 9 cos^{2} (θ) + sin (θ) + 11 = - sin (θ) + 2

What is the general solution for 9cos(2x)=10sin(x)+1 ?
The general solution for 9cos(2x)=10sin(x)+1 is x=(3pi)/2+2pin,x=0.46055…+2pin,x=pi-0.46055…+2pin