find exactly all solutions to the equations?

Question

a) cos 2(theta) + cos(theta) = 0
-for theta is greater than or equal to 0 and less than 360 degrees

b) 2cos^2(theta) = 3sine(theta) + 3
-for theta is greater than or equal to 0 and less than or equal to 2pi

Answer 1

you should try to rewrite each equation in terms of either just cos or sin. I'll do the first one:

use the double angle formula
cos 2(theta) = cos^2(theta) - sin^2(theta)
= 2cos^2(theta) - 1,

so your equation
cos 2(theta) + cos(theta) = 0
becomes
2cos^2(theta) - 1 + cos(theta) = 0.

now if you make the substitution x = cos(theta), this equation is a quadratic:
2x^2 - 1 + x = 0,
and you can factor it as
(2x - 1) (x + 1) = 0,
so x = 1/2 or -1.

now you just have to solve
cos(theta) = 1/2 or -1.
cos(theta) = 1/2 if theta = 60 or 300 degrees,
and cos(theta) = -1 if theta = 180 degrees.

Answer 2

cos theta = -3/5

Answer 3

Let x = theta
cos 2x + cos x = 0
2cos^2x -1 +cos x = 0
cos x = [-1+/-sqrt(1+ 4*2)]/4
cos x = [-1+/- 3]/4
cosx = -1/4 +/- 3/4 = -1 or 1/2
So x = 180, 60, 300 degrees

2cos^x =3sinx +3
2(1-sin^2x) -3sinx - 3 = 0
-2sin^2x -3sinx -1 = 0
sinx = [3 +/-sqrt(3^2 - 4(-2) (-1)]/-4
sinx = [ 3 +/- 1]/-4
sinx = -1 or -1/2
x = 270, 330, 210 degrees

Answer 4

you can do that by checking the ratio of the coefficients of variables in the eqns. if the ratios are unequal then the eqns have a single soln. ex:5x+6y+7z=3,3x+2y+4z=7,x+y+z=8 If the ratios of the coefficients are equal then you have to check the ratio of values of each eqn. if the ratios of values are not equal to the ratio of coefficients,then the eqns have no soln. Ex:2x+3y+4z=5,4x+6y+8z=3,6x+9y+12z=8 But if the ratio of values of eqns are equal to the ratio of the coefficients,then the eqns have infinite soln. ex:2x+3y+4z=5,4x+6y+8z=10,6x+9y+12z=15