Question

Pre-Cal

Answer 1

Solve the following equation for 0° ≤ θ < 360°. Use the "^" key on the keyboard to indicate an exponent. For example, sin2x would be typed as sin^2x. Be sure to show all your work.

2sin^2x - cos^2 x - 2 = 0

Answer 2

hint first change the cos^2x to sin

remember the magic identity sin^2x _+ cos^2x = 1 well
1 - sin^2x = cos^2x start with this substitution

Answer 3

Okay, I am very new to this. So as I understand it now, I should have

2sin^2x-sin-2=0

Answer 4

after putting in 1 -sin^2 you have
\[ 2\sin^2x - (1 - \sin^2x ) -2 =0\]

distribute the -1:
\[ 2\sin^2x - 1 + \sin^2x -2 =0\]

collect "like terms" :
\[ 3\sin^2x -3 =0\]

Answer 5

we don't literally "change cos^2 to sin"
we use the identity
sin^2 x + cos^2 x = 1 (this is well-known)
from which we know
cos^2 x = 1 - sin^2 x

and we use that to replace the cos^2

Answer 6

so then it would be 2sin^2x-1-sin^2x=0

Answer 7

\[3(\sin ^2 -1)\]=0

Answer 8

divide both sides by 3

and then factoring you have difference of 2 squres
(sinx-1)(sinx+1) = 0

set both these to zero and solve
sinx-1=0

Answer 9

sinx+1 = 0

Answer 10

So is that the answer or are there more steps?

Answer 11

@zpupster

Answer 12

sinx-1=0
sinx=1 and

sinx+1 = 0
sinx=-1

find where sinx = 1 and -1