Question

using double angle identities, find the exact solutions for 0 11 years ago

Answer 1

Are you sure it is 5 cos2 in RHS ? @joshnel

Answer 2

yes

Answer 3

NO sorry, it is 5cosx RHS

Answer 4

Yeah thought so :)

Answer 5

\[\large{3+\cos 2x = 5\cos x}\]

First of all we will use this formula:

\[\large{\cos 2x = 2\cos^2 x - 1}\]

Answer 6

We will get:

\[\large{3+2\cos^2 x - 1 = 5\cos x}\]

\[\large{\implies 2\cos^2 x + 2 - 5\cos x = 0}\]

Answer 7

Any doubt ?

Answer 8

No, and now we factor?

Answer 9

Yeah

Answer 10

\[\large{2\cos^2 x - 4\cos x - \cos x + 2 = 0}\]

\[\large{\implies 2\cos x (\cos x - 2) - (\cos x -2 ) = 0}\]

\[\large{\implies (2\cos x - 1)(\cos x - 2) = 0}\]

Answer 11

then set each factor to zero, then substituts cos=y, solve for y, then sub y=cos back into the equation?

Answer 12

Perfect !!!!!!!!!

Just remember that cos x cannot be less than -1 and more than 1

Answer 13

Suppose that it is? or is that just impossible?

Answer 14

\[\large{-1 \le \cos x \le 1}\]

Thus, one solution would automatically be discarded as cos x = 2 is not a possible value

Answer 15

wait, so the factor (cosx-2) is automatically disregarded?

Answer 16

Yup... you will have to consider only 2cos x - 1 = 0

Answer 17

Because cos x - 2= 0 will give cos x = 2 which is not possible

Answer 18

Oh, i see, i literally just repeated what you typed, sorry about that.

Answer 19

Its okay

Answer 20

so the only possible solution is cos = 1/2

Answer 21

Yup

Answer 22

ok, thank you so much, i really appreciate the help. I think i have it down, its just applying the double angle formulas

Answer 23

Your welcome :)