# Thread: [SOLVED] trig function cos^2

1. ## [SOLVED] trig function cos^2

Solve: $\displaystyle 4 cos^2 x = 3$
for
0 < x < 2pi

I thought the way I did it was right but I get the wrong answer.

so I factored
cosx(4cosx)=3
so
cosx=3 and 4cosx=3
cosx cannot =3 so 4cosx=3 ....
cosx=3/4

so I sketch a little graph and find x=pi/6 and 15 pi/8? but this gets confusing. whats the right way to find this?
but that isnt right, could someone help me with this question?

thanks.

2. Originally Posted by brentwoodbc
Solve: $\displaystyle 4 cos^2 x = 3$
for 0 < x < 2pi
No not quite.
$\displaystyle 4 cos^2(x)=3$
$\displaystyle cos(x)=\pm \frac{\sqrt{3}}{2}$

3. Originally Posted by Plato
No not quite.
$\displaystyle 4 cos^2(x)=3$
$\displaystyle cos(x)=\pm \frac{\sqrt{3}}{2}$
thanks I just realized that.
but my teacher said you could factor and solve like.
$\displaystyle cos^2x-1=0$
cos(x)=0 and cosx=1
x=0,pi/2,3pi/2,2pi ? is that wrong?

4. Originally Posted by brentwoodbc
my teacher said you could factor and solve like. $\displaystyle cos^2x-1=0$
cos(x)=0 and cosx=1
x=0,pi/2,3pi/2,2pi ? is that wrong?
Yes it is wrong.
$\displaystyle 4cos^2(x)-3=0$
$\displaystyle cos^2(x)-\frac{3}{4}=0$
$\displaystyle (cos(x)-\frac{\sqrt3}{2})(cos(x)+\frac{\sqrt3}{2})=0$

5. sorry an example of the way I did it is
sin^2x-sinx=0
sinx = 0 and sinx-2=0

this is one of the ways he taught us, he said if its a quadratic to do this, isnt the first example I gave you a quadratic with the b term =0 so why doesnt it work?

6. Originally Posted by Plato
Yes it is wrong.
$\displaystyle 4cos^2(x)-3=0$
$\displaystyle cos^2(x)-\frac{3}{4}=0$
$\displaystyle (cos(x)-\frac{\sqrt3}{2})(cos(x)+\frac{\sqrt3}{2})=0$
thanks that answers my last post.

Thank you.