# 3 Trig Equations and 1 Word Problem Help

• May 3rd 2009, 08:32 PM
College Student
3 Trig Equations and 1 Word Problem Help
1. 2sin(x+(pi/2)) + 3tan(pi-x) = 0, in the interval [0, pi)

2. cos2x + 2sin^2(x) - 2tan^2(x) - 1 = 0, in the interval [0, 2pi)

3. 4sin^2(x) - 4sin(x) + 1 = 0, in the interval [0, 2pi)

4. Two airplanes leave an airport at the same time, one going west at 375 mph and the other going northeast at 425 mph. How far are they apart two hours after departure?

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This is how far I got for problem #1:

Step 1: 2(sinxcos(pi/2)+cosxsin(pi/2) + 3(sin(pi-x)/cos(pi-x)) = 0
Step 2: 2(sinxcos(pi/2)+cosxsin(pi/2) + 3(sinpicosx-cospisinx/cospicosx+sinx) = 0

I don't know where to go from there and I don't know how to get started on the other two equations. And for the word problem, I thought about using the Law of Cosines, but I need an angle to use that formula. I don't think this is a right triangle problem either because of the second plane going NE. Please help.
• May 3rd 2009, 09:25 PM
pickslides
Quote:

Originally Posted by College Student
1. 2sin(x+(pi/2)) + 3tan(pi-x) = 0, in the interval [0, pi)

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This is how far I got for problem #1:

Step 1: 2(sinxcos(pi/2)+cosxsin(pi/2) + 3(sin(pi-x)/cos(pi-x)) = 0
Step 2: 2(sinxcos(pi/2)+cosxsin(pi/2) + 3(sinpicosx-cospisinx/cospicosx+sinx) = 0

After using the identities in this link

Table of Trigonometric Identities

I arrived at sin(x) = -1/2 which has no solution on the interval [0,pi)
• May 3rd 2009, 09:49 PM
Soroban
Hello, College Student!

Here's the first one . . .

Quote:

$1)\;\;2\sin\left(x+\tfrac{\pi}{2}\right) + 3\tan(\pi-x) \:=\: 0$ in the interval $[0, \pi)$
Simplfy first . . .

. . $\sin\left(x + \tfrac{\pi}{2}\right) \:=\:\sin x\cos\tfrac{\pi}{2} + \cos x \sin\tfrac{\pi}{2} \;=\;\sin x\cdot0 + \cos x\cdot 1 \;=\;\cos x$

. . $\tan(\pi - x) \:=\:\frac{\tan\pi -\tan x}{1 + \tan\pi\tan x} \;=\;\frac{0-\tan x}{1 - 0} \;=\;-\tan x$

The problem becomes: . $2\cos x - 3\tan x \:=\:0 \quad\Rightarrow\quad 2\cos x - 3\frac{\sin x}{\cos x} \:=\:0$

. . $2\cos^2\!x - 3\sin x \:=\:0 \quad\Rightarrow\quad 2(1-\sin^2\!x) - 3\sin x \:=\:0$

We have: . $2\sin^2\!x + 3\sin x - 2 \:=\:0 \quad\Rightarrow\quad (\sin x + 2)(2\sin x - 1)\:=\:0$

. . $\sin x + 2 \:=\: 0 \quad\Rightarrow\quad\sin x \:=\: -2 \quad\text{no solution}$

. . $2\sin x - 1 \:=\: 0 \quad\Rightarrow\quad \sin x \:=\: \tfrac{1}{2} \quad\Rightarrow\quad x \:=\: \tfrac{\pi}{6},\:\tfrac{5\pi}{6}$

• May 3rd 2009, 10:56 PM
College Student
Thanks, Soroban.

Here's what I got so far for the second problem:

Step 1: 1-2sin^2(x) + 2 sin^2(x) - 2(sinx/cosx) - 1 = 0
the bolded parts gets cancelled out (I think)
Step 2: -2(sinx/cosx) = 0

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That's about as far as I got. #3 still confuses me and I still have no luck with the word problem.

Edit: Oh, wait. Instead of converting tan to sin and cos, if I kept it like it is:

-2tan^2(x) = 0
tan^2(x) = 0
tanx = 0
x = 0, pi, 2pi

Is that right?