Expert: Paul Klarreich - 5/9/2006

Question
I need to solve these three in terms of x

Cos((5pi)/2)+x=-Sin(x)

Cos(x)> -radical(3)/2

Sin(2x)<0

Thank You

Answer
Hi, Lee,
 
Subject:  I need help as soon as possible
Question:  I need to solve these three in terms of x
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Please, in the future, the subject should relate to the SUBJECT MATTER, not to your rush.  I always answer as fast as I can, whether you are just curious or desperate.

These are questions about SOLVING TRIGONOMETRIC EQUATIONS.
Unfortunately, they are not always routine and they test your knowledge of trigonometric properties and identities.

A.  Cos((5pi)/2)+x=-Sin(x)

I think you mean:

Cos((5pi)/2)+x) = - Sin(x)

Use a cos(A+B) expansion:  cos(A+B) = cosA cosB - sinA sinB

cos 5pi/2 cos x - sin 5pi/2 sin x = - sin x

But 5pi/2 = 2pi + pi/2, so
cos 5pi/2 = cos pi/2 = 0
sin 5pi/2 = sin pi/2 = 1

0 cos x - 1 sin x = - sin x
       - sin x = - sinx
So your equation is an identity.  That means it is true for all x.
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B. Cos(x)  >= - sqrt(3)/2

For this, you make these observations:
1. If it said Cos(theta) = - sqrt(3)/2   { not >= }
then you would proceed as follows:

cos theta = sqrt(3)/2  has the solution  theta = 30 degrees, or pi/6, in the first quadrant.  But you have   cos(theta) = -sqrt(3)/2, so this has to be in quadrants 2 and 3.  In Q2, the solution is 150 degrees, or 5pi/6.  In Q3, the solution is 210 degrees, or 7pi/6.

2. But it does say  cos theta >= -sqrt(3)/2.  So what values of theta will satisfy it?  Obviously, only values between  5pi/6 and 7pi/6 FAIL to, because the x-coordinates actually go further to the left.  So we can conclude:

theta = 0 to 5pi/6, + 2n pi and
theta = 7pi/6 to 2pi, + 2n pi.

This is not so easy to explain.  I suggest you have your calculator draw the graph of y = cos x from 0 to 2pi, and then have it draw the line y = -sqrt(3)/2.   You will see that very little of the graph actually goes below the line; most of it does satisfy  cos x >= -sqrt(3)/2


C. Sin(2x) < 0

There are a couple of ways to do this.  Easiest, I think is:

sin T < 0 means  T is between pi and 2pi.  (In quadrants 3 and 4)

So that says    pi + 2npi < T < 2pi + 2npi
Then:

pi + 2npi < 2x < 2pi + 2npi

Divide by 2:

pi/2 + npi < x < pi + npi

What does that look like?

x can be between  pi/2 and  pi, meaning Quadrant 2, and
x can be between 3pi/2 and 2pi, meaning Quadrant 4.

OR, here's another way: you can write:
sin(2x) = 2 sin x cos x, so

2 sin x cos x < 0
 sin x cos x < 0

Now that means  sin x and cos x have opposite signs, if their product is negative.  Where is that?  In Q1, both are positive -- no good.  In Q2, sine is +, cosine is -,  GOOD.  In Q3, both are negative -- no good.  Q4 is OK, too.