Solve the equation for all values of x.
-2cos²x-sin x+1=0, on the interval [0,2π).
-(2cosx-1)(cosx+1) = 0
so, where is cosx = 1/2 or -1?
To solve the equation -2cos²x - sin x + 1 = 0, we need to find the values of x that satisfy this equation on the interval [0, 2π). Here's how you can solve it step by step:
Step 1: Rewrite the equation in terms of cos(x) and sin(x)
Since the equation includes cos²x and sin x, let's rewrite it using the trigonometric identity: sin²x + cos²x = 1. Replace cos²x with 1 - sin²x in the equation:
-2(1 - sin²x) - sin x + 1 = 0
Step 2: Simplify the equation
Distribute -2 to both terms inside the parentheses:
-2 + 2sin²x - sin x + 1 = 0
Combine like terms:
2sin²x - sin x - 1 = 0
Step 3: Factor the equation
We can factor this quadratic equation:
(2sin x + 1)(sin x - 1) = 0
Step 4: Set each factor equal to zero and solve separately
Setting 2sin x + 1 = 0:
2sin x = -1
sin x = -1/2
Setting sin x - 1 = 0:
sin x = 1
Step 5: Solve for x
To find the values of x, we use the inverse sine function, sin⁻¹, and solve separately for both cases:
For sin x = -1/2:
In the interval [0, 2π), the solutions are x = 7π/6 and x = 11π/6.
For sin x = 1:
In the interval [0, 2π), the solution is x = π/2.
Therefore, the values of x that satisfy the equation on the interval [0, 2π) are x = 7π/6, 11π/6, and π/2.