3cos(2x)+4sin(2x)=2

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Solution for 3cos(2x)+4sin(2x)=2 equation: 


Simplifying
3cos(2x) + 4sin(2x) = 2

Remove parenthesis around (2x)
3cos * 2x + 4sin(2x) = 2

Reorder the terms for easier multiplication:
3 * 2cos * x + 4sin(2x) = 2

Multiply 3 * 2
6cos * x + 4sin(2x) = 2

Multiply cos * x
6cosx + 4sin(2x) = 2

Remove parenthesis around (2x)
6cosx + 4ins * 2x = 2

Reorder the terms for easier multiplication:
6cosx + 4 * 2ins * x = 2

Multiply 4 * 2
6cosx + 8ins * x = 2

Multiply ins * x
6cosx + 8insx = 2

Solving
6cosx + 8insx = 2

Solving for variable 'c'.

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

Add '-8insx' to each side of the equation.
6cosx + 8insx + -8insx = 2 + -8insx

Combine like terms: 8insx + -8insx = 0
6cosx + 0 = 2 + -8insx
6cosx = 2 + -8insx

Divide each side by '6osx'.
c = 0.3333333333o-1s-1x-1 + -1.333333333ino-1

Simplifying
c = 0.3333333333o-1s-1x-1 + -1.333333333ino-1

Reorder the terms:
c = -1.333333333ino-1 + 0.3333333333o-1s-1x-1

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