3sin(a)+4cos(b)=8

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Solution for 3sin(a)+4cos(b)=8 equation:


Simplifying
3sin(a) + 4cos(b) = 8

Multiply ins * a
3ains + 4cos(b) = 8

Multiply cos * b
3ains + 4bcos = 8

Solving
3ains + 4bcos = 8

Solving for variable 'a'.

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

Add '-4bcos' to each side of the equation.
3ains + 4bcos + -4bcos = 8 + -4bcos

Combine like terms: 4bcos + -4bcos = 0
3ains + 0 = 8 + -4bcos
3ains = 8 + -4bcos

Divide each side by '3ins'.
a = 2.666666667i-1n-1s-1 + -1.333333333bci-1n-1o

Simplifying
a = 2.666666667i-1n-1s-1 + -1.333333333bci-1n-1o

Reorder the terms:
a = -1.333333333bci-1n-1o + 2.666666667i-1n-1s-1

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