Please Help
Solve each equation for 0<x<2 pi
2cos^ 2 x +sins-1=0
4 answers
Please proof read.
Solve equation for 0< Equality x < 2pi
Trig equations
Trig equations
I'm not sure if this is right but this is what I have so far
2cos^2 +sinx -1 =0
2cos ^2 x+sinx =1 (I added 1 to both sides)
Cos^2 x +sinx
2cos^2 +sinx -1 =0
2cos ^2 x+sinx =1 (I added 1 to both sides)
Cos^2 x +sinx
you have to keep writing an equation; you just dropped off the right side. In any case, remember from your Algebra I that you usually need to set stuff to zero to solve.
Now, using the fact that sin^2x+cos^2x=1,
2cos^2x+sinx-1 = 0
2-2sin^2x + sinx-1 = 0
2sin^2x-sinx-1 = 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1
Now list the angles in all the quadrants that fit.
Now, using the fact that sin^2x+cos^2x=1,
2cos^2x+sinx-1 = 0
2-2sin^2x + sinx-1 = 0
2sin^2x-sinx-1 = 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1
Now list the angles in all the quadrants that fit.
2 answers