2cos^2+sin=1
To solve the equation 2cos^2(x) + sin(x) = 1, we can use some trigonometric identities and algebraic techniques.
1. First, we know that cos^2(x) = 1 - sin^2(x), so we can substitute this into the equation:
2(1 - sin^2(x)) + sin(x) = 1
2. Distribute the 2 to both terms inside the parenthesis:
2 - 2sin^2(x) + sin(x) = 1
3. Rearrange the equation to isolate sin(x) terms on one side:
-2sin^2(x) + sin(x) - 1 = 0
4. This equation is a quadratic equation in terms of sin(x). We can solve it by factoring or using the quadratic formula.
Factoring:
The quadratic equation -2sin^2(x) + sin(x) - 1 = 0 can be factored as:
(-2sin(x) + 1)(sin(x) + 1) = 0
Set each factor to zero:
-2sin(x) + 1 = 0 or sin(x) + 1 = 0
Solving these equations separately:
-2sin(x) = -1 or sin(x) = -1
sin(x) = 1/2 or sin(x) = -1
5. To find the values of x that satisfy sin(x) = 1/2 or sin(x) = -1, we can refer to the unit circle or trigonometric identities.
For sin(x) = 1/2:
One solution is x = π/6 (corresponding to 30 degrees) as sin(π/6) = 1/2.
Another solution is x = 5π/6 (150 degrees) since sin(5π/6) = 1/2.
For sin(x) = -1:
One solution is x = 3π/2 (270 degrees) because sin(3π/2) = -1.
Therefore, the solutions to the equation 2cos^2(x) + sin(x) = 1 are:
x = π/6, 3π/2, and 5π/6.