Question

I have a Pre-Calc Test Monday, and our teacher told us one of the questions is sinx^2+cosx^2+1=0 or sinx^2+cosx+1=0. SOLVE FOR GENERAL SOLUTIONS, please! I am not sure which one is the correct problem, so could someone help me solve both? Thanks so much!

Answer 1

is that
\[ \left(\sin(x)\right)^2 = \sin^2(x)\]
or \[ \sin\left(x^2\right) \]

Answer 2

It is the first set (sin(x))2=sin2(x)

Answer 3

in that case, you should remember this useful fact
\[ \sin^2(x) + \cos^2(x) = 1 \]

Answer 4

if you use that identity in the first equation you gave, you get 2=0
which is not true.

Answer 5

you can use it in the second equation you have, to get a quadratic in cos(x)

Answer 6

Wouldn't the 1 be -1 in the original equation if I was using the identity sin2(x)+cos2(x)=1 because it would have been moved over to the other side?

Answer 7

\[ \sin^2x+\cos^2x+1=0\]
use the identity to replace the first 2 terms with 1. you get
1+1 = 0

Answer 8

Oh Okay! I see now!

Answer 9

for the 2nd one
\[ \sin^2x+\cos x+1=0 \\ (1 -\cos^2 x ) + \cos x+1=0 \]
that simplifies to
\[ -\cos^2 x + \cos x +2 = 0 \\ \cos^2 x - \cos x - 2 = 0 \]
if you let y = cos x, this is
\[ y^2 -y -2 = 0 \]
factor to get
\[ (y-2)(y+1) = 0 \]
or
(y-2) =0 --> cos x = 2 Not possible (cos goes -1 to +1)
y+1=0 --> cos x = -1
now find all angles x where cos x is -1

Answer 10

Thank you so much!!!!!!