Sin(2B+5)=cos(3B-5)

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Solution for Sin(2B+5)=cos(3B-5) equation: 


Simplifying
Sin(2B + 5) = cos(3B + -5)

Reorder the terms:
inS(5 + 2B) = cos(3B + -5)
(5 * inS + 2B * inS) = cos(3B + -5)

Reorder the terms:
(2inBS + 5inS) = cos(3B + -5)
(2inBS + 5inS) = cos(3B + -5)

Reorder the terms:
2inBS + 5inS = cos(-5 + 3B)
2inBS + 5inS = (-5 * cos + 3B * cos)
2inBS + 5inS = (-5cos + 3cosB)

Solving
2inBS + 5inS = -5cos + 3cosB

Solving for variable 'i'.

Move all terms containing i to the left, all other terms to the right.

Reorder the terms:
5cos + -3cosB + 2inBS + 5inS = -5cos + 3cosB + 5cos + -3cosB

Reorder the terms:
5cos + -3cosB + 2inBS + 5inS = -5cos + 5cos + 3cosB + -3cosB

Combine like terms: -5cos + 5cos = 0
5cos + -3cosB + 2inBS + 5inS = 0 + 3cosB + -3cosB
5cos + -3cosB + 2inBS + 5inS = 3cosB + -3cosB

Combine like terms: 3cosB + -3cosB = 0
5cos + -3cosB + 2inBS + 5inS = 0

The solution to this equation could not be determined.

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