How do you solve #2\cos^{2} x + \cos 2x + \sin x = 0#?

Question

Oliver Atkinson · Accepted Answer

We will need the following identities to solve this equation. #•cos^2x = 1 - sin^2x#
#•cos2x = 1 - 2sin^2x# Our goal here is to change everything into sine.  Here it goes: #2(1 - sin^2x) + 1 - 2sin^2x + sinx = 0# #2 - 2sin^2x + 1 - 2sin^2x + sinx = 0# #0  = 4sin^2x - sinx - 3# Let #t = sinx#. #0 = 4t^2 - t - 3# #0 = 4t^2 - 4t + 3t - 3# #0 = 4t(t - 1) + 3(t-  1)# #0 = (4t + 3)(t - 1)# #t = -3/4 and 1# #sinx = -3/4 and sinx = 1# #x = pi + arcsin(3/4), 2pi - arcsin(3/4), pi/2# Hopefully this helps!

Josiah Anderson · Answer

undefined To solve $2\cos^{2} x + \cos 2x + \sin x = 0$, follow these steps:

1. Rewrite $\cos 2x$ using the double angle identity: $\cos 2x = 2\cos^{2} x - 1$.

2. Substitute the rewritten expression into the equation: $2\cos^{2} x + (2\cos^{2} x - 1) + \sin x = 0$.

3. Combine like terms: $4\cos^{2} x - 1 + \sin x = 0$.

4. Rearrange the equation: $4\cos^{2} x + \sin x - 1 = 0$.

5. Recognize that this equation can be transformed into a quadratic equation by substituting $u = \cos x$: $4u^{2} + \sin x - 1 = 0$.

6. Since $\sin x = \sqrt{1 - \cos^{2} x}$, substitute $\sin x$ in terms of $\cos x$: $4u^{2} + \sqrt{1 - u^{2}} - 1 = 0$.

7. Now, solve the quadratic equation for $u$, and then find the corresponding values of $x$.