4 cos 2x + 3 sin 2x = 2
expanding things a bit, we have
4cos^2x - 4sin^2x + 6sinx cosx = 2
2cos^2x + 3sinx cosx - 2sin^2x = 1
(2cosx-sinx)(2sinx+cosx) = 1
Now divide by cos^2x and you have
(2-tanx)(2tanx+1) = sec^2x
4tanx - 2tan^2x + 2 - tanx = 1 + tan^2x
3tan^2x - 3tanx - 1 = 0
tanx = (3±√21)/6
Now you can get your angles from that.
I assume we are solving ....
4√(1- sin^2 2x) ) + 3sin 2x = 2
√(1-sin^2 2x) = (2 - 3sin 2x)
/4
square both sides
1 - sin^2 2x = (4 - 12sin 2x + 9sin^2 2x)/16
let sin 2x = y, so we have
1 - y^2 = (4 - 12y + 9y^2)/16
16 - 16y^2 = 4 - 12y + 9y^2
25y^2 - 12y - 12 = 0
y = (12 ±√1344)/50
= .973212111 or y = -.49321211
case 1:
sin2x = .9732...
2x = 1.33881 or 2x = π - 1.33881... = 1.8027...
x = .6694 or x = .90139
case 2:
sin 2x = -.493212..
2x = π + .515778 or 2x = 2π - .515778
x = 1.8287 or x = 2.8837
Now , since I squared the equation, all answers MUST be verified.
subbing in my values of x, the two that worked are
x = .90139 and x = 2.8837
since the period of both sin 2x and cos 2x is π
adding or subtracting multiple of π will yield more answers
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.