Question

Precalculus

Answer 1

\[2 \sin^2 x + 3 \cos x - 3 = 0\]

@Hero

Solving for x when [0, 2pi)

Answer 2

\(\sin^2x + \cos^2x = 1\)
\(\sin^2x = 1 - \cos^2x\)

Answer 3

\[2(1 - \cos^2 x ) + 3 \cos x - 3 = 0\]
\[2 - 2\cos ^2 x + 3 \cos x - 3 = 0\]
\[-2\cos ^2 x + 3 \cos x - 1 = 0\]

Where do I go from here?

Answer 4

\(y = \cos(x)\)
\(-2y^2 +3y - 1 = 0\)

Answer 5

@Shadow

Answer 6

\[(-2\cos x + 1)(\cos x - 1) = 0\]

Answer 7

\[(-2 \cos x + 1) = 0\]
\[\cos x - 1 = 0\]

Answer 8

\[x = 0, \frac{ \pi }{ 3 }, \frac{ 5\pi }{ 3 }\]

Answer 9

Did I miss any angles?

Answer 10

Graph it and see. Use desmos.com and set the increments and window limitations in terms of \(\pi\)

Answer 11

It's just those three angles, cause we're staying within 0 and 2pi

Answer 12

Did you graph it on desmos? If so, post it here.

Answer 13

Don't know how to use desmos. Put it in wolfram and three angles

Answer 14

Ah I see. If Wolfram says it's right, then it is correct. Graphing is always a good way to verify if an answer is correct.

Answer 15

Ah, I see. Okay