Please Help Me Solve for Theta (I will use x to make it easier)?

Question

2cos^2(x)+cos (x)=0

tan (x)=2sin (x)

Answer 1

cos x(2 cos x + 1) = 0  (factor out a cos x)

either cos x = 0 ==> x = pi/2 or 3pi/2 or

2 cos x + 1 = 0 ==> cos x = -1/2 ==> x = 2pi/3 or 4pi/3

solutions between 0 and 2pi: x = pi/2, 2pi/3, 4pi/3, 3pi/2

general solutions:

x = pi/2 + 2pi*k

x = 2pi/3 + 2pi * k

x = 4pi/3 + 2pi * k

x = 3pi/2 + 2pi * k

where k is an integer... the 2pi*k factor adds or subtracts multiples of the period of cos x to get back to the same values of cos

tan x = 2 sin x

sin x / cos x = 2 sin x (use identity that tan = sin / cos)

sin x = 2 sin x cos x (multiply both sides by cos x)

2 sin x cos x - sin x = 0 (move sin x to the other side)

sin x (2 cos x - 1) = 0 (factor out a sin x)

so either sin x = 0 ==> x = 0, pi, 2pi

or 2cos x - 1 = 0 ==> cos x = 1/2 ==> x = pi/3 or 5pi/3

solutions between 0 and 2pi:  0, pi/3, pi, 5pi/3, 2pi

general solution: 0 + pi*k

pi/3 + 2pi * k

5pi/3 + 2pi * k

Answer 2

cosx(2cosx+1) = 0

cos x = 0 --> x pi/2 or 3pi/2

cos x = -1/2 --> x = 2pi/3 or 4pi/3

sinx/cosx = 2sinx

1/cosx = 2

cosx = 1/2

x = pi/3 or 5pi/3

Answer 3

From equation 1,

(cos x)(2cos x +1)=0

cos x can be 0 which gives x as 180n+/-90.(Here / stands for or and n is an integer)

or cos x = -1/2

which gives us the values of x as 180(2n+1)+/-60

From equation 2,

sin x/cos x=2sin x

So sin x can be 0 which gives x=180n

or cos x=1/2

which gives us x=180(2n)+/-60

so combining all the ranges of x we get no common values.

Hence there is no x satisfying both the equations.

Answer 4

This requires either a little bit of cleverness and remembering your trig identities, or else recognizing that the first equation is just a quadratic equation.  (2x^2 + x = 0, or x(2x+1) = 0), which gives two possible values of x (i.e. cos(theta)=0 or -1/2), or 2 possible values of cos(theta), and therefore 4 possible values of theta (remember why?).  That should be enough of a hint, the rest you can get yourself.

Answer 5

1. Let n = cos x.  Then 2n^2 = -n or -2n^2 = n, and n = -1/2.  Then cos x = -.5 and x = 135 degrees.

2. sin x/cos x = 2sin x.  Dividing both sides by 2sin x and multiplying both sides by cos x yields cos x = 1/2 and x = 60 degrees.