# Pre Cal

posted by .

2cos*2x-cosx=1 find all solutions to the equation on the domain [0,2ð]

i got ð/4, 7ð/4, 3ð/4, and 5ð/4 but only got two out of four points... what did i do wrong?

• Pre Cal -

Is cos*2x supposed to mean cos^2 x?
Is 2ð supposed to mean 2 pi?
If so,
2 cos^2x - cos x -1 = 0
which factors into
(2 cos x +1)(cos x -1) = 0
The solutions occur where cos x = -1/2 or +1

If cos*2x means cos(2x), then
2[2 cos^2 x -1) - cos x -1 = 0
4 cos^2 x -cos x -3 = 0
(4 cosx +3)(cosx - 1) = 0
x is where cos x = -3/4 or +1

## Similar Questions

1. ### math,correction

Find four solutions for the equation 3x+5y=15 so the equation will turn to be y=-(3)/(5)x+ (3) and when i do the table i get these points for the solution (-2,4.2),(-1,3.60),(0,3),(1,1.8) ok Let me give you a help. Take the slope, …
2. ### Trig.......

I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr Reiny …
3. ### pre cal--check

Name all the values of x that are not in the domain of f(x) = (2–x^2)/ (x+ 5) I got only x= -5 because that's the only value that would make the denominator zero.
4. ### algebra

Will someone please look over these? I am so lost on these. 1. find the domain of p(x)=xsquared+10- I said {x|x is a real number 2. find the intercepts of 3x+2y=6- I got (0,3) and (2,0) 4.find the equation of (-6,-6)and (6,6)- got
5. ### Pre-Cal

Find all the solutions of the equation in the interval (0,2pip) sin(x + pi/6) - sin(x -pi/6) = 1/2 I am only stuck on the last part. I have 2sinx(sqrt3/2) = 1/2 Does the 2 cancel out with the sqrt 3/2. I am not sure what to do.
6. ### Pre-Cal(Urgent!!)

Find all the solutions of the equation in the interval (0,2pip) sin(x + pi/6) - sin(x -pi/6) = 1/2 I am only stuck on the last part. I have 2sinx(sqrt3/2) = 1/2 Does the 2 cancel out with the sqrt 3/2. I am not sure what to do.