Question

Solve the equation for x

0<= x <=2pi

sin^2x - sqrt(2)cosx = cos^2x + sqrt(2)cosx + 2

Answer 1

i got 0=cos^2x + sqrt(2)cosx + 1


I tried using the quadratic formula but i get a discriminate with a negative.

Answer 2

help?

Answer 3

\[\sin ^2x-\sqrt{2}\cos x=\cos ^2x+\sqrt{2}\cos x+1\]

Answer 4

Is that the problem or is there parentheses around x+1 ?

Answer 5

yes, tht is the problem except it is + 2 not + 1

Answer 6

\[\sin^2x - \sqrt{2}cosx = \cos^2x + \sqrt{2}cosx + 2\]

Answer 7

\(\color{red}{\sin^2x} - \sqrt2 \cos x = \cos^2x + \sqrt2 \cos x + 2\)

Use the identity \(\sin^2 x = 1 - \cos^2 x\)

\( \color{red}{ 1 - \cos^2 x } - \sqrt2 \cos x = \cos^2x + \sqrt2 \cos x + 2\)

\(2 \cos^2 x + 2 \sqrt{2} \cos x + 1 = 0 \)

Answer 8

Now use the quadratic formula.

Answer 9

Use the quadratic formula to solve for \( \cos x\) and then use trig to solve for x.

Answer 10

wow, thank you guys so much!

Answer 11

wlcm