To solve the equation 4cos(2x) + 3sin(2x) = 2, we can use trigonometric identities to simplify it.
First, recall the double angle formulas:
cos(2x) = cos^2(x) - sin^2(x)
sin(2x) = 2sin(x)cos(x)
Substituting these values into the equation, we get:
4(cos^2(x) - sin^2(x)) + 3(2sin(x)cos(x)) = 2
Expanding and rearranging the equation, we have:
4cos^2(x) - 4sin^2(x) + 6sin(x)cos(x) = 2
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:
4cos^2(x) - 4(1 - cos^2(x)) + 6sin(x)cos(x) = 2
Expanding further, we have:
4cos^2(x) - 4 + 4cos^2(x) + 6sin(x)cos(x) = 2
Combining like terms, we get:
8cos^2(x) + 6sin(x)cos(x) - 2 = 0
Now we can solve this quadratic equation in terms of cos(x). We introduce a substitution to simplify the equation: let u = cos(x).
Substituting u back into the equation, we get:
8u^2 + 6sin(x)u - 2 = 0
This is now a quadratic equation in terms of u. We can use the quadratic formula to solve for u:
u = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 8, b = 6sin(x), and c = -2. Substitute these values into the quadratic formula to find the value(s) of u.
After obtaining the solutions for u, substitute u back into the equation u = cos(x) to find the solutions for cos(x).
Finally, since cos(x) = u, solve for x by taking the inverse cosine (cos^(-1)) of each value of u to get the values of x.
Note that there may be multiple solutions for x, depending on the values of u obtained from the quadratic equation.