What are all the solutions of cos 2x + cos x - 2 = 0?

Question

a 0º, 360º

b 0º + 360nº

c 0º, 180º, 360º

d 0º + 180nº

Answer 1

cos 2x + cos x - 2 =0

2cos^2 x -1 + cos x -2 =0

Let a=cos x. Then,

2a^2 + a - 3 =0

(2a + 3)(a - 1) =0

a=-3/2 or a=1

Since a=cos x

and cosx>-1

cosx=1

x=0 + 180n

For the second question, plot sinx and (1/810)x in the same graph.

I think the answer would be five or nine.

Answer 2

First Problem:

2cos^2x+cosx -3 =0

2cos^2x +3cosx - 2cosx -3=0

cosx[2cosx+3]-[2cosx+3]=0

[cosx-1][2cosx+3]=0

cosx=1 or -3/2 inwhich -3/2 is not possible

Hence, cosx=1

==> x= 0 degree or 360 degrees or multiples of 360 degrees

So, the answer is [b]

Second Problem:

As sinx ranges from -1 to 1, x/810 must range from -1 to 1

by converting into radians, pi*x/[180*810] must range from -1 to 1

Hence, x must range from -46433.12 to 46433.12

So, the number of revolutions is more than nine. if you take in each revolution, the answer may exceed nine

Answer 3

2 cos ² x - 1 + cos x - 2 = 0

2 cos ² x + cos x - 3 = 0

(2 cos x + 3)(cos x - 1) = 0

cos x = 1 is acceptable

x = 0° + 360n°

OPTION b