2cos(2x)+2((cos^2)x)=1-3sinx

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Solution for 2cos(2x)+2((cos^2)x)=1-3sinx equation:


Simplifying
2cos(2x) + 2((cos2) * x) = 1 + -3sinx

Remove parenthesis around (2x)
2cos * 2x + 2((cos2) * x) = 1 + -3sinx

Reorder the terms for easier multiplication:
2 * 2cos * x + 2((cos2) * x) = 1 + -3sinx

Multiply 2 * 2
4cos * x + 2((cos2) * x) = 1 + -3sinx

Multiply cos * x
4cosx + 2((cos2) * x) = 1 + -3sinx

Multiply cos2 * x
4cosx + 2(cos2x) = 1 + -3sinx

Solving
4cosx + 2cos2x = 1 + -3insx

Solving for variable 'c'.

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

Reorder the terms:
-1 + 4cosx + 2cos2x + 3insx = 1 + -3insx + -1 + 3insx

Reorder the terms:
-1 + 4cosx + 2cos2x + 3insx = 1 + -1 + -3insx + 3insx

Combine like terms: 1 + -1 = 0
-1 + 4cosx + 2cos2x + 3insx = 0 + -3insx + 3insx
-1 + 4cosx + 2cos2x + 3insx = -3insx + 3insx

Combine like terms: -3insx + 3insx = 0
-1 + 4cosx + 2cos2x + 3insx = 0

The solution to this equation could not be determined.

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