Question

find the exact solutions of the equation in the interval [0,2pi) cos 2x + sin x = 0

Answer 1

You need to first change that double angle to an identity that involves a sin or a sin squared or something. The cos(2x) double angle has three different identities. One of those is very fitting to our needs, considering we will have to factor this to find the solutions. The identities for cos(2x) are as follows:

Answer 2

\[\cos(2x)=\cos ^{2}x-\sin ^{2}x\]\[\cos(2x)=1-2\sin ^{2}x\]\[\cos(2x)=2\cos ^{2}x-1\]

Answer 3

The goal is to get all the functions the same. in other words, either pick sin or cos, but not both. Since we already have the sin function in our original, and cos(2x) has a couple sin functions in our identities, let's change the cox(2x) to a sin function, preferable one that involves a sin^2 so we can factor it. I'd pick this one:\[\cos(2x)=1-2\sin ^{2}x\]

Answer 4

When we do that we end up with this: (stay with me...)

Answer 5

\[1-\sin ^{2}x+\sin x=0\]

Answer 6

We could rearrange that to read like this:

Answer 7

\[-2\sin ^{2}x+\sin x +1=0\]

Answer 8

Then change the sign within the equation to get rid of the negative out front:

Answer 9

\[2\sin ^{2}x-\sin x -1=0\]

Answer 10

Not for 100%, but I'm guessing that that is something that we can factor. If it helps, replace the sin(x) with u and factor u^2 - u - 1 = 0.

Answer 11

Are you with me?

Answer 12

Solve this trig equation, in the form of a quadratic equation, with sin x = t:
2t^2 - t - 1 = 0. This is the case of TIP 1: a + b + c = 0. One real root is 1 and the other is c/a = -1/2. Next, solve the 2 basic trig equations:
1. sin x = t = 1= sin Pi/2 --> x = Pi/2
2. sin x = t = -1/2 = sin 4Pi/3 -> x = 4Pi/3 ; x = 5Pi/3

Answer 13

Sorry for the mistake:
2. sin x = t = -1/2 = sin 7Pi/6 = sin 11Pi/6 --> x = 7Pi/6 ; x = 11/Pi/6