# Solve: http://postimage.org/image/beuqg9s2f/Identities and Equations

lfryerda | High School Teacher | (Level 2) Educator

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Only the first question may be annswered.  You may put your second question as a separate question.

Solve the equation for 0<=x<=360^\circ.

6cos^2 theta+cos theta - 1=0

This is a quadratic equation in cos theta, which can be factored to get

(2cos theta+1)(3cos theta - 1)=0

The first factor on the LHS gives the equation

2cos theta+1=0

which is the same as cos theta =-1/2

By the CAST rule and special triangles, we know that cos 60^\circ =1/2 , so the two solutions where cosine are negative are in the second and third quadrants.

This means that two solutions are theta=180-60=120^\circ and theta=180+60=240^circ.

For the second factor, there is no special triangle, so we need to approximate the solution using a calculator.  The second factor gives the equation cos theta = 1/3, which has a has approximate solution theta = 70.5^\circ.  From the CAST rule, we know that the other solution to this equation is in the fourth quadrant, so the two solutions are \theta=70.5^\circ and \theta = 360-70.5=289.5^circ.