Question

Find all solution in the interval [0,2pi) by algebraically: 2sin^2 2x+5sin2x-3=0

Answer 1

Let \(u=\sin2x\) and see our equation reduces to a (reducible) quadratic, \(2u^2+5u-3=0\). Can you factor it?

Answer 2

Not sure how to...

Answer 3

Look for factors of \(2\times-3=-6\) that sum to \(5\). Then we split our middle term and factor by groupinh.

Answer 4

Thanks for trying to help out, but I give up. This just isn't my forte.

Answer 5

It's just factoring a polynomial... algebra, not trigonometry.$$2u^2+5u-3=0\\2u^2+6u-u-3=0\\2u(u+3)-1(u+3)=0\\(2u-1)(u+3)=0$$... so it follows that either \(2u-1=0\) or \(u+3=0\). Solve:$$2u-1=0\\2u=1\\u=\frac12$$or$$u+3=0\\u=-3$$

Answer 6

Now recall that \(u=\sin 2x\), so to find \(x\) we now direct our attention to \(\sin2x=\frac12\) or \(\sin2x=-3\) (not possible ;-)

Answer 7

And then?

Answer 8

Can you solve \(\sin 2x=\frac12\)? What angles give you \(\frac12\) for sine?

Answer 9

pi/6 and 5pi/6 right?

Answer 10

@Mathgirl503 those angles do indeed, but try expanding your horizons to include the other two results below \(4\pi\). These are our solutions in \(2x\), so we divide by \(2\) to yield solutions in \(x\).