SOLUTION: Instructions: Solve. Give exact answers in radians, 0<_x<_2pi Problems : 1. 2cos^2 x + 3cosx - 2 = 0 2. tan^2 x - secx =1 Thank you so much (: I'm just co

Algebra ->  Trigonometry-basics -> SOLUTION: Instructions: Solve. Give exact answers in radians, 0<_x<_2pi Problems : 1. 2cos^2 x + 3cosx - 2 = 0 2. tan^2 x - secx =1 Thank you so much (: I'm just co      Log On


   

Question 1090341: Instructions: Solve. Give exact answers in radians, 0<_x<_2pi
Problems : 1. 2cos^2 x + 3cosx - 2 = 0
2. tan^2 x - secx =1
Thank you so much (: I'm just completely lost with these two problems and have no idea how to start
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!
These are both quadratic equations with some trig function as the "variable". Solve them by getting 0 on one side of the equation and factoring, exactly as you would do for an ordinary quadratic equation.
The first one is in the required form for factoring:
2cos%5E2%28x%29+%2B+3cos%28x%29+-+2+=+0
%282cos%28x%29-1%29%28cos%28x%29%2B2%29+=+0
The first factor gives you cos(x) = 1/2, so x is pi/3 or 5pi/3.
The second factor gives you cos(x) = -2, for which there is no solution.
So pi/3 and 5pi/3 are the solutions to the first one.
For the second one, you have some work to do to get it in the right form. And the first thing you want to do is get everything in terms of sine and cosine, since those are easier to work with. Then you will want to use appropriate identities to get the equation in terms of either sine or cosine alone.
tan%5E2%28x%29+-+sec%28x%29+=+1
convert to sines and cosines...
sin%5E2%28x%29%2Fcos%5E2%28x%29+-+1%2Fcos%28x%29+=+1
clear fractions by multiplying by cos^2(x)...
sin%5E2%28x%29+-+cos%28x%29+=+cos%5E2%28x%29
convert the sin^2(x) term into an expression using cosines...
%281-cos%5E2%28x%29%29+-+cos%28x%29+=+cos%5E2%28x%29
get everything on one side, set equal to 0...
2cos%5E2%28x%29+%2B+cos%28x%29+-+1+=+0
Now factor and solve as in the other example.