Question

sin^2 theta + cos theta = 2

(Hint: Use the Pythagorean identity sin^2 theta + cost^2 theta = 1 to replace sin^2 theta in the given equation.)

Answer 1

\[ \sin^2 \theta + \cos^2 \theta = 1 \\ \sin^2 \theta = 1-\cos^2 \theta\]

Answer 2

\[1-\cos^2\Theta = 2\]

So then it would be this^?

Answer 3

you start with
\[ \sin^2 \theta + \cos \theta = 2 \]
replace the sin^2 with 1 - cos^2
in that equation. what do you get ?

Answer 4

\[1-\cos^2\Theta+cosTheta=2\]

Answer 5

you get
\[ 1 - \cos^2 \theta + \cos \theta = 2\]
which simplifies to
\[ - \cos^2 \theta + \cos \theta -1 =0 \\
\cos^2 \theta - \cos \theta +1 =0 \]

if you let x= cosθ
you have
\[ x^2 -x +1 = 0\]
you will find x= (1± sqr(3) i) /2
this is an imaginary number. That means there is no real solution.

So you are doing a complicated problem or you have a typo in the problem

Answer 6

Yeah, maybe there's a typo x)

Answer 7

Thanks for your help!