Thread: unsure+ confused!?

Solve 2sin^2 -7sin = -3 for , where 0°< < 360°.

How would I do this???

Originally Posted by superman

Solve 2sin^2 -7sin = -3 for , where 0°< < 360°.

How would I do this???

2 sin^2(theta) - 7 sin(theat) = -3

is a quadratic in sin(theta). Solve this quadratic using to quadratic formula
to get sin(theta) = 1/2 or 3. Now sin(theta)=3 has no solution for theta in
the given range, so we are left with:

sin(theta) = 1/2, for theta in (0, 360] degrees, which you should be able to do

RonL

Hello, superman!

Solve: .2·sin²θ -7·sin θ .= .-3 .for 0°< θ < 360°.

We have a quadratic equation: .2·sin²θ - 7·sin θ + 3 .= .0

. . which factors: .(sin θ - 3)(2·sin θ - 1) .= .0


And we have two equations to solve:

. . sin θ - 3 .= .0 . → . sin θ .= .3 . . . . no real solutions

. . 2·sin θ - 1 .= .0 . → . sin θ = ½ . → . θ .= .30°, 150°