write the general solution to 2cos^2x+cos x=1
it factors...
2cos^2x + cosx - 1 = 0
(2cosx - 1)(cosx + 1) = 0
cosx = 1/2 or cosx = -1
x = 60° or 300° or x = 180°
general solution..... add 360k° to each answer
I will leave it up to you to change the above to radians if you have to.
Treat cosx as a new variable u, and solve the resulting quadratic equation for u.
2 u^2 +u -1 = 0
(2u -1)(u + 1) = u
u = cos x = -1 or 1/2
Now solve for x. One answer is pi radians. Another is pi/3. There is another, in the fourth quadrant.
To find the general solution to the equation 2cos^2x + cos x = 1, we can follow these steps:
Step 1: Rewrite the equation in terms of cosine squared.
2cos^2x + cos x = 1 [Substitute cos^2x = 1 - sin^2x]
2(1 - sin^2x) + cos x = 1 [Distribute 2]
2 - 2sin^2x + cos x = 1 [Rearrange terms]
2 - 2sin^2x + cos x - 1 = 0 [Combine like terms]
2 - 2sin^2x + cos x - 1 = 0 [Combine like terms]
Step 2: Simplify the equation.
-2sin^2x + cos x + 1 = 0 [Combine like terms]
Step 3: Rearrange the equation.
-2sin^2x + cos x + 1 = 0 [Rearrange terms]
2sin^2x - cos x - 1 = 0 [Multiply by -1]
Step 4: Factor the equation.
(2sin x + 1)(sin x - 1) = 0 [Factor the equation]
Step 5: Set each factor equal to zero and solve for x.
2sin x + 1 = 0 [Set first factor equal to zero]
sin x = -1/2 [Solve for sin x]
sin x - 1 = 0 [Set second factor equal to zero]
sin x = 1 [Solve for sin x]
Step 6: Solve for x.
For sin x = -1/2, we can use the unit circle or trigonometric identity to find the values of x. The solutions are x = 7π/6 + 2πn, x = 11π/6 + 2πn, where n is an integer.
For sin x = 1, x = π/2 + 2πn, where n is an integer.
Therefore, the general solution to the equation 2cos^2x + cos x = 1 is:
x = 7π/6 + 2πn, x = 11π/6 + 2πn, x = π/2 + 2πn, where n is an integer.