# Question #dcf23

##### 1 Answer

#### Explanation:

We have the equation

#6cos^2theta-costheta=1#

Notice that this is similar to a quadratic equation--we have a variable expression being squared, a variable expression to the first power, and a constant. We should try to solve this as we normally would a quadratic first and *then* bring the trigonometry into play.

To make this more approachable, a good (although not necessary method) is to let

#6u^2-u-1=0#

To factor this quadratic, find a pair of factors of

#6u^2-3u+2u-1=0#

#3u(2u-1)+1(2u-1)=0#

#(3u+1)(2u-1)=0#

#{(3u+1=0),(2u-1=0):}#

#{(u=-1/3),(u=1/2):}#

Now, since

#{(costheta=-1/3),(costheta=1/2):}#

Solving just for the first equation:

#costheta=-1/3#

Using the inverse cosine function,

#theta=arccos(-1/3)#

Plugging this into a calculator yields

The second equation is a commonly known value:

#costheta=1/2#

#theta=60˚#