Question

How do I solve these trigonometric equations!!?

Find all solutions in the interval [0, 2π).

(sin x)(cos x) = 0

Find all solutions in the interval [0, 2π).

cos2x + 2 cos x + 1 = 0

Find all solutions to the equation.

cos2x + 2 cos x + 1 = 0

(I am not looking for easy answers to cheat, I just need help on how to solve them...Im a bit confused and my teacher is not avail)

Answer 1

A unit circle might be helpful. Let's do the first together.
(sinx)(cosx)=0 implies that sinx=0 or cosx=0
when does sinx=0? when does cosx=0?

Answer 2

are you familiar with the unit circle?

Answer 3

I'm very sorry my internet went out.

I am familiar with the unit circle...

So sinx = 0 at 90 degrees? and coax = 0 at 270 degrees?

Answer 4

Wait, correction... sinx equals 0 at 0 degrees and cosx = 0 at 90 degrees
right?

Answer 5

Sinx =0 at 0 and also pi and multples of pi. Cosx =0 at pi/2 and odd multiples of pi/2. There you go.

Answer 6

Ok, now how do I go about solving the equations?

Answer 7

First, for clarification is that second one cos2x as you have written or is it supposed to be cos^2x?

Answer 8

That is cos^2x

Answer 9

Well, that makes our job very simple. \[\cos ^{2}x+2\cos x+1=\left( \cos x+1 \right)^{2}\]

So if we set that = 0, you should be able to find x fairly easily.

Answer 10

Hm would that be x = pi/2 and 3pi/2 ?

Answer 11

There you go. Now either the last one is a repeat, or this time it really is a cos2x.

Answer 12

The second one is the same, however the only difference is the question it asks, for the second one it asks find all the solutions in the interval (0, 2pi) and the second one asks to find all solutions...

Answer 13

I mean last one*

Answer 14

you should re-think
(cos x +1 )^2 =0

Answer 15

Wait, neither of those values work.
l

Answer 16

Phi is correct. The answers are not multiples of pi/2.

Answer 17

There is only one angle within the period of cos that works.

Answer 18

I am a bit confused.