What is the general solution for sin(b-20)=cos(142.7) ?

The general solution for sin(b-20)=cos(142.7) is b=360n-32.7,b=180+360n+72.7

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sin(b-20)=cos(142.7)

sin (b - 2 0^{\circ}) = cos (142. 7^{\circ})

b = 36 0^{\circ} n - 32. 7^{\circ}, b = 18 0^{\circ} + 36 0^{\circ} n + 72. 7^{\circ}

Solution steps

sin (b - 2 0^{\circ}) = cos (142. 7^{\circ})

Rewrite using trig identities

sin (b - 2 0^{\circ}) = sin (9 0^{\circ} - 142. 7^{\circ})

Apply trig inverse properties

b - 2 0^{\circ} = 9 0^{\circ} - 142. 7^{\circ} + 36 0^{\circ} n, b - 2 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - 142. 7^{\circ}) + 36 0^{\circ} n

b - 2 0^{\circ} = 9 0^{\circ} - 142. 7^{\circ} + 36 0^{\circ} n : b = 36 0^{\circ} n - 32. 7^{\circ}

b - 2 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - 142. 7^{\circ}) + 36 0^{\circ} n : b = 18 0^{\circ} + 36 0^{\circ} n + 72. 7^{\circ}

b = 36 0^{\circ} n - 32. 7^{\circ}, b = 18 0^{\circ} + 36 0^{\circ} n + 72. 7^{\circ}

2 \cdot (1 - cos^{2} (x)) - cos^{2} (x) = 1 + sin^{2} (x)

cos (θ) = \frac{11}{61}

1 - cos^{2} (x) = 3 sin (x) - 2

cos (x) = \frac{4}{8}

\frac{sin ( 5 1 ^{\circ} )}{18} = \frac{sin ( x )}{22}

What is the general solution for sin(b-20)=cos(142.7) ?
The general solution for sin(b-20)=cos(142.7) is b=360n-32.7,b=180+360n+72.7