# Trigonometric Equations?

• May 10th 2009, 12:14 PM
Frontier
Trigonometric Equations?
I have an assignment here on trig equations but I'm not sure how to do them. I have a vague idea but my textbook doesn't offer much help and I won't be able to ask my teacher for assistance until Thursday. Would anybody be willing to explain how they work please?

1) Determine the exact value in simplest form for sin (-135 degrees)/2cos(420 degs) - Sin (240 degs).

I'm just not sure where to begin with this. I thought I'd be able to use the unit circle but it isn't working out.

2) Determine all solutions for each equation:

a) 4cos(3x - 80) + 5.5 = 4.3

b) 8sin^2x+ 10sinx - 3 = 0

What does it mean by "all solutions"?

I thought I'd be able to work it like a normal equation, so I changed A to make 4cos(3x-80) = -1. Then I put my reference angles in quadrants 2 and 3 of the unit circle, got ref angle = cos^-1 (1) and couldn't do anything else because my calculator can't solve it. What'd I do wrong? It's the same story with B; I work it as far as the ref angle and my calculator tells me I'm doing it wrong.

3) Determine the x intercepts of the function f(x) = 6sin [3/4 (x-120)] + 5 on the interval xE[-360, 360].

That one completely goes over my head. I'm pretty sure I'm supposed to create a graph for it, but I dont' even know where to begin. What does it want? And what does "on the interval xE[-360, 360]" mean?
• May 10th 2009, 01:18 PM
JoanF
Quote:

Originally Posted by Frontier
I have an assignment here on trig equations but I'm not sure how to do them. I have a vague idea but my textbook doesn't offer much help and I won't be able to ask my teacher for assistance until Thursday. Would anybody be willing to explain how they work please?

1) Determine the exact value in simplest form for sin (-135 degrees)/2cos(420 degs) - Sin (240 degs).

I'm just not sure where to begin with this. I thought I'd be able to use the unit circle but it isn't working out.

2) Determine all solutions for each equation:

a) 4cos(3x - 80) + 5.5 = 4.3

b) 8sin^2x+ 10sinx - 3 = 0

What does it mean by "all solutions"?

I thought I'd be able to work it like a normal equation, so I changed A to make 4cos(3x-80) = -1. Then I put my reference angles in quadrants 2 and 3 of the unit circle, got ref angle = cos^-1 (1) and couldn't do anything else because my calculator can't solve it. What'd I do wrong? It's the same story with B; I work it as far as the ref angle and my calculator tells me I'm doing it wrong.

3) Determine the x intercepts of the function f(x) = 6sin [3/4 (x-120)] + 5 on the interval xE[-360, 360].

That one completely goes over my head. I'm pretty sure I'm supposed to create a graph for it, but I dont' even know where to begin. What does it want? And what does "on the interval xE[-360, 360]" mean?

You need to think in the unit circle.

1) think: -135º= -180º+45º, so we can conclude that the sin(-135º) = sin(-180º+45º) = - sin 45º = - square of 2 / 2

do the same to the others and substitute in the equation for the results

2) all solutions means that you have to solve the exercice in |R and not in a certain interval (for example, the interval [0, pi])

3) I think that here you have to see where the function intercepts x-axis, so you do:
f(x) = 0
6sin [3/4 (x-120)] + 5=0
sin [3/4 (x-120)] =-5/6
then you check, on the calc, where the sine is -5/6,
sin alfa = -5/6
alfa= -56,44º

then:
sin [3/4 (x-120)]=sin(-56,44º)
3/4(x-120)=-56,44 + 360k or 3/4(x-120)=180-(-56,44)+360k, kEZ
x= 44,75+480k or x=435,25+480k, kEZ and xE[-360, 360]

then you check the solutions of the equation on that interval giving values to k:

for k=0, x=44,75 or x=435,25 (44,75 belongs to the given interval but 435,25 not, so 44,75 is one solution)
for k=-1, x=-435,25 or x=-44,75 (-44,75 belongs to the given interval but -435,25 not, so -44,75 is one solution)

so the solutions are: -44,75 and 44,75
• May 10th 2009, 01:38 PM
Frontier
Quote:

1)
Quote:

think: -135º= -180º+45º, so we can conclude that the sin(-135º) = sin(-180º+45º) = - sin 45º = - square of 2 / 2
Ohhhh, okay. So I just get the neg value of what Sin135 is on the Unit Circle? So square 2/2 would = -square 2/2, alright. Thanks.

So for 2cos(420)-Sin(240), would it be:

2cos(120 + 300)+square 3/2?

What would I do with the 2cos(120+300)? Just 2(-1/2 + square 3/2)?
• May 10th 2009, 01:50 PM
JoanF
Quote:

Originally Posted by Frontier
[b]

Ohhhh, okay. So I just get the neg value of what Sin135 is on the Unit Circle? So square 2/2 would = -square 2/2, alright. Thanks.

So for 2cos(420)-Sin(240), would it be:

2cos(120 + 300)+square 3/2?

What would I do with the 2cos(120+300)? Just 2(-1/2 + square 3/2)?

No you have the check on the unit circle with values that you know the sin, cos and tg (they are: 0º,30º,45º,60º,90º,180º an 360º) you have to comparate the angles. For example: 135º, you can think that 135=90º+45º, then you draw the angles 135º and 45º and you comparate the sin, cos or tg. You want to know the sin 135, you draw this angle in the unit circle and you compare sin135º with sin45º and you see that they are equals.
http://www.postimage.org/aV1l3_gr.jpg
• May 10th 2009, 02:45 PM
Frontier
Ohh, okay. I understand. Thank you very much! : )