Trig question?

Question

Solve for x in the interval 0 < or equal to x < 360:

2 cos x - 1 = 1 / cos x

Work and/or explanation, please!

Answer 1

can't have cos x = 0 else denominator blows up (so x != 90, 270). Also, 0 <= x < 360.

Multiply through by cos(x):
2 cos²(x) - cos(x) = 1

Move everything over to one side:
2 cos²(x) - cos(x) - 1 = 0

Factor:
(2cos(x) + 1)(cos(x) - 1) = 0 so

2cos(x) = -1 or cos(x) = 1
Second one says x = 0 (You've ruled out 360)

First one says cos(x) = -1/2
or x = 180 ± 60 => 120, 240

Solutions are therfore:
0, 120, 240 degrees

Answer 2

Notice that cos(x) is in the denominator of the right. This means that cos(x) cannot be zero, i.e., x cannot be 90 or 270.

Now, multiply through by cos(x) to clear the fractions:
2 cos(x) ^2 - cos(x) = 1.

Take all to the left:
2 cos(x)^2 - cos(x) - 1 = 0.

Factor:
(2 cos(x) + 1)(cos(x) - 1) = 0.

So, cos(x) = -1/2 or cos(x) = 1.

cos(x) = -1/2 when x = 150 and when x = 240.

cos(x) = 1 when x = 0.

x = 0, 150, 240.

Answer 3

Remember that cos x cannot be zero ....
Multiply both side by it.


2 cos^2 x - cos x = 1

For convenience let t=cos x

2t^2 - t - 1 = 0

t1,2 = [1 +- sqrt(1 + 4*2*1)]/2 = [1 +- sqrt(9)]/2 = (1 +- 3)/4

t1 = (1 + 3)/4 = 1.

t2 = (1 - 3)/4 = -2/4 = -1/2

If cos x = 1 x=0.

If cos x=-1/2 x=120 or 240

Answer 4

2 cos x - 1 = 1 / cos x
2cos²x -cosx = 1
2cos²x -cosx -1 =0
Δ = (-1)² -4(2)(-1) = 1+8 = 9
√Δ = 3

cosx =(+1±3)/4
cosx = (1+3)/2 = 2 it's not possible cos' -1≤ cosx ≤+1
it rests
cosx =(1-3)/4 = -1/2
angles where cosx it's equal (-1/2):
2π/3 and 4π/3

Besides of:
if x = 0º => cos0º =1
2.cos0 -1 = 1/cos0
1=1 ( T )
x=2π
Solution:
S={ 2π, 2π/3, 4π/3 }

that's all!!!