Question

Can someone help me, please?

Answer 1

Solve each trig EQUATION for x, where 0< x <or equal to 2pi:
sin^2x-3cosx=3

Answer 2

x = Pi

Answer 3

How did you figure that out?

Answer 4

@jim_thompson5910

Answer 5

Refer to the Wolfram solution attached.

Answer 6

It says the answer is pi+2pi?

Answer 7

@robtobey

Answer 8

HI!!

Answer 9

\[\sin^2(x)-3\cos(x)-3=0\] is a start
then replace \(\sin^2(x)\) by \(1-\cos^2(x)\) and get a quadratic equation in cosine

Answer 10

you get
\[1-\cos^2(x)-3\cos(x)-3=0\] or
\[\cos^2(x)+3\cos(x)+2=0\] then factor it

Answer 11

\[(\cos(x)+2)(\cos(x)+1)=0\] so \(cos(x)=-2\) which is not possible or \(\cos(x)=-1\) which is possible if \(x=\pi\)

Answer 12

Thank you!!