Question

Solve the following equation
sin2 x = -2cos x + 2

Answer 1

that is sin(2x) right?

Answer 2

on the left hand side ?

Answer 3

its sin^2(x)

Answer 4

\[\sin^2(x)=-2\cos(x)+2 \text{ write \in terms of just } \cos \\ 1-\cos^2(x)=-2\cos(x)+2\]

Answer 5

this is a quadratic in terms of the function cos(x)

Answer 6

put everything on one side

Answer 7

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

Answer 8

you can factor the right hand side

Answer 9

do you know how to factor u^2-2u+1?

Answer 10

nope. no idea how to do any of this

Answer 11

hint (a-b)^2=a^2-2ab+b^2
so u^2-2u+1 =?

Answer 12

if you really forgot how to factor quadratics
you could use the quadratic formula

Answer 13

if \[a \cdot u^2+b \cdot u+c=0 \\ \text{ then } u=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 14

can you identify from your equation what a,b, and c are?

Answer 15

are b and c 2?

Answer 16

you have
\[(\cos(x))^2-2(\cos(x))+1=0 \\ \text{ you are comparing it to } \\ a(u)^2+b(u)+c=0\]

Answer 17

what is the coefficient in front of (cos(x))^2
what is the coefficient in front of the (cos(x))
what is the constant term?

Answer 18

-2
1
0

Answer 19

hmm so that means you are looking at a totally different equation
like -2cos^2(x)+cos(x)+0=0