Question

Solve for x: sin^2x = 4-2cos^2x

Answer 1

\(\large\color{black}{ \sin^2x = 4-2\cos^2x }\)
there is a rule, \(\large\color{red}{ \sin^2x+\cos^2x=1 }\) and when we subtract \(\large\color{red}{ \cos^2x }\) from both sides, we get \(\large\color{red}{ \sin^2x=1 -\cos^2x }\)

So,
\(\large\color{black}{ (1-\cos^2x )= 4-2\cos^2x }\)

Answer 2

(have re-written sin^2x in terms of cos^2x. )

Answer 3

Tell me if you need more help from this point and on.

Answer 4

so the cos should be sqrt(3)?

Answer 5

Let's see....
\(\large\color{black}{ 1-\cos^2x =4-2\cos^2x }\)
adding cos^2x to both sides,
\(\large\color{black}{ 1 =4-3\cos^2x }\)
subtract 4 from both sides
\(\large\color{black}{ -3 =-3\cos^2x }\)

Answer 6

can you take it from here?

Answer 7

i got \[\cos ^{2} x = 3\]
therefore \[\cos x = \sqrt{3}\]

Answer 8

after the last thing I wrote, I proceed by dividing by -3 on both sides, I get,
\(\large\color{black}{ \cos^2x=1 }\)
then I take a square root of both sides,
\(\large\color{black}{ \cos x=\pm1 }\)

\(\large\color{blue }{ x=1,~~\pi }\)

Answer 9

but -2cos^2x + cos^2 x = -cos^2 x

Answer 10

yes....

Answer 11

and so?

Answer 12

lets see i think you are correct @KRJones23

Answer 13

@SolomonZelman , when you added cos^2 .... it should have been -2+1 = -1

Answer 14

ohh, true lol.

Answer 15

so then 3=cos^2x.

Answer 16

cos x must be within -1 and 1 so this equation has no solution

Answer 17

yuup. (shouldn't be saying that, because sounds like I am the one who "knows", but still yes)