What is the general solution for 9sin(b)+3=5sin(b)+1 ?

The general solution for 9sin(b)+3=5sin(b)+1 is b=(7pi)/6+2pin,b=(11pi)/6+2pin

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9sin(b)+3=5sin(b)+1

9 sin (b) + 3 = 5 sin (b) + 1

b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

Solution steps

9 sin (b) + 3 = 5 sin (b) + 1

Solve by substitution

sin (b) = - \frac{1}{2}

sin (b) = - \frac{1}{2} : b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

Combine all the solutions

b = \frac{7 π}{6} + 2 πn, b = \frac{11 π}{6} + 2 πn

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

cos (x) = 2 cos^{2} (x) - 1

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

3 tan (2 x + 1) = 2

\frac{1}{( 1 + tan ( a ) )} = cot (a)

What is the general solution for 9sin(b)+3=5sin(b)+1 ?
The general solution for 9sin(b)+3=5sin(b)+1 is b=(7pi)/6+2pin,b=(11pi)/6+2pin