I need help solving for all solutions for this problem:
cos 2x+ sin x= 0
I substituted cos 2x for cos^2x-sin^2x
So it became cos^2(x)-sin^2(x) +sinx=0
Then i did 1-sin^2(x)-sin^2(x)+sinx=0
= 1-2sin^2(x)+sinx=0
= sinx(-2sinx+1)=-1
What did i do wrong??
the real solutions are 7pi/6, 11pi/6, and pi/2
2 sin^2 - sin - 1 = 0
(2 sin + 1)(sin - 1) = 0
2 sin + 1 = 0 ... sin = -1/2
sin - 1 = 0 ... sin = 1
To solve the equation cos 2x + sin x = 0, let's break it down step by step.
1. Start with the equation: cos 2x + sin x = 0.
2. Use the double angle formula for cosine: cos 2x = 2cos^2 x - 1. Substitute this into the equation: 2cos^2 x - 1 + sin x = 0.
3. Rearrange the equation: 2cos^2 x + sin x - 1 = 0. Now, we have a quadratic-like equation.
4. Notice that we can factorize this equation: (2cos x - 1)(cos x + 1) = 0.
5. Set each factor equal to zero and solve for x:
- 2cos x - 1 = 0, which simplifies to cos x = 1/2, giving x = π/3 + 2πn, where n is an integer.
- cos x + 1 = 0, which simplifies to cos x = -1, giving x = π + 2πn, where n is an integer.
These solutions give us values for x where the equation is satisfied, but we need to check if they work in the original equation.
For the first case:
- x = π/3 + 2πn. Substituting this into the original equation, we get: cos 2(π/3 + 2πn) + sin(π/3 + 2πn) = 0.
- Simplify using trigonometric identities and periodicity: cos(4π/3 + 4πn) + sin(π/3 + 2πn) = cos π/3 + sin π/3 = 1/2 + √3/2 = 1 ≠ 0.
This means that the solutions for x = π/3 + 2πn do not satisfy the original equation.
For the second case:
- x = π + 2πn. Substituting this into the original equation, we get: cos 2(π + 2πn) + sin(π + 2πn) = 0.
- Simplify using trigonometric identities and periodicity: cos(2π + 4πn) + sin(π + 2πn) = cos 0 + sin π = 1 + 0 = 1 ≠ 0.
This means that the solutions for x = π + 2πn do not satisfy the original equation.
Therefore, there are no real solutions for the provided equation cos 2x + sin x = 0.
If you were expecting different solutions (such as 7π/6, 11π/6, and π/2), please double-check the given equation or provide any additional information.