What is the general solution for sin(x+b)=cos(x) ?

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

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sin(x+b)=cos(x)

sin (x + b) = cos (x)

x = - 2 πn - \frac{b}{2} + \frac{π}{4}, x = - \frac{3 π}{4} - 2 πn - \frac{b}{2}

Solution steps

sin (x + b) = cos (x)

Subtract

cos (x)

from both sides

sin (x + b) - cos (x) = 0

Rewrite using trig identities

- 2 sin (\frac{π - 2 b}{4}) sin (\frac{π - 2 b - 4 x}{4}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin (\frac{π - 2 b - 4 x}{4}) = 0

General solutions for

sin (\frac{π - 2 b - 4 x}{4}) = 0

\frac{π - 2 b - 4 x}{4} = 0 + 2 πn, \frac{π - 2 b - 4 x}{4} = π + 2 πn

Solve

\frac{π - 2 b - 4 x}{4} = 0 + 2 πn : x = - 2 πn - \frac{b}{2} + \frac{π}{4}

Solve

\frac{π - 2 b - 4 x}{4} = π + 2 πn : x = - \frac{3 π}{4} - 2 πn - \frac{b}{2}

x = - 2 πn - \frac{b}{2} + \frac{π}{4}, x = - \frac{3 π}{4} - 2 πn - \frac{b}{2}

3 sin (ϕ) - cos (2 ϕ) = 1

2 tan^{2} (θ) + 3 tan (θ) + 4 = 3

10 cos (x) = 8

2 sin^{2} (x) = 6 cos^{2} (x)

1 + cot^{2} (θ) = 1.33

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