Solve 2cos^2(x) + 4sin(x)-3 = 0, giving all solutions in the interval 0 ≤ x ≤ 2π
I already did this today
180-30
360 - 30
270 in degrees
multiply degrees by pi/180 to get radians if that is what you need
http://www.jiskha.com/display.cgi?id=1447870446
oh, you changed a number
2 (1 -sin^2) + 4 sin - 3 = 0
2 - 2 sin^2 + 4 sin - 3 = 0
-2 sin^2 + 4 sin - 1 = 0
2 sin^2 - 4 sin + 1 = 0
(2 sin -1)(2 sin-1) = 0
sin = 1/2
angle = 30 or 150 degrees
To solve the equation 2cos^2(x) + 4sin(x) - 3 = 0 in the interval 0 ≤ x ≤ 2π, we can use the trigonometric identities and techniques.
Step 1: Simplify the equation by using the identity cos^2(x) + sin^2(x) = 1.
- The equation becomes 2(1 - sin^2(x)) + 4sin(x) - 3 = 0.
- Distribute the 2: 2 - 2sin^2(x) + 4sin(x) - 3 = 0.
- Combine like terms: -2sin^2(x) + 4sin(x) - 1 = 0.
Step 2: Divide the entire equation by -1 to make the coefficient of the leading term positive:
-(-2sin^2(x) + 4sin(x) - 1) = 0.
2sin^2(x) - 4sin(x) + 1 = 0.
Step 3: Factor or use the quadratic formula to solve the quadratic equation.
2sin^2(x) - 4sin(x) + 1 = (2sin(x) - 1)(sin(x) - 1) = 0.
Set each factor equal to zero and solve for sin(x):
2sin(x) - 1 = 0 --> sin(x) = 1/2
sin(x) - 1 = 0 --> sin(x) = 1
Step 4: Find all solutions in the interval 0 ≤ x ≤ 2π using the unit circle or calculator:
For sin(x) = 1/2, there are two angles that satisfy this condition: x = π/6 and x = 5π/6.
For sin(x) = 1, there is one angle that satisfies this condition: x = π/2.
So, the solutions in the interval 0 ≤ x ≤ 2π are x = π/6, 5π/6, and π/2.