What is the general solution for sin(b)=cos(2b-30) ?

The general solution for sin(b)=cos(2b-30) is b=(pi+60+4pin)/6 ,b=-(pi-60+4pin)/2

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sin(b)=cos(2b-30)

sin (b) = cos (2 b - 30)

b = \frac{π + 60 + 4 πn}{6}, b = - \frac{π - 60 + 4 πn}{2}

Solution steps

sin (b) = cos (2 b - 30)

Rewrite using trig identities

sin (b) = sin (\frac{π}{2} - (2 b - 30))

Apply trig inverse properties

b = \frac{π}{2} - (2 b - 30) + 2 πn, b = π - (\frac{π}{2} - (2 b - 30)) + 2 πn

b = \frac{π}{2} - (2 b - 30) + 2 πn : b = \frac{π + 60 + 4 πn}{6}

b = π - (\frac{π}{2} - (2 b - 30)) + 2 πn : b = - \frac{π - 60 + 4 πn}{2}

b = \frac{π + 60 + 4 πn}{6}, b = - \frac{π - 60 + 4 πn}{2}

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

\frac{sin ( \frac{π}{2} )}{5} = \frac{sin ( x )}{4}

cos (3 x - 1) = \frac{1}{2}

tan^{2} (x) = sec (x) + 1

cos (u) = \frac{2}{5}, cos (2 u)

What is the general solution for sin(b)=cos(2b-30) ?
The general solution for sin(b)=cos(2b-30) is b=(pi+60+4pin)/6 ,b=-(pi-60+4pin)/2