Due Tomorrow! Solving Trig equations

• Mar 15th 2007, 10:45 AM
leviathanwave
Due Tomorrow! Solving Trig equations
Hey, I'm stuck on these 5 problems and can't seem to figure out what's what. I'm supposed to express in terms of *blah* + 2k(pi) i believe. Any help appreciated. I'll continue working on them and check back later. I really need the points for these, so I kinda need to get em all right ^_^

1. Find all solutions of the following. Express in radians.

a. 2cos^2(x-3)cos(x+1) = 0

b. 4tan(3x - 4) = 0

c. sin(2x - 1) = -1/2

2. Find the exact solutions of sin(x+2)csc(x) = 3 on the interval [0, 2(pi)]. Express in radians.

3. Find all solutions of csc(x) + cot(x) = sqrt(3). express answer in radians. Check answers. Hint: rewrite in terms of sine and cosine, clear fractions by multiplying through by the denominator, and square both sides.
• Mar 15th 2007, 01:32 PM
Soroban
Hello, leviathanwave!

Here's some help . . .

Quote:

1. Find all solutions of the following. .Express in radians.

a. .2·cos²(x - 3)·cos(x + 1) .= .0

Set each factor equal to zero.

2·cos²(x - 3) = 0 . . cos(x - 3) = 0 . . x - 3 .= .½π + 2kπ . . x .= .½π + 3 + 2kπ

cos(x + 1) = 0 . . x + 1 .= .±½π + 2kπ . . x .= . ±½π - 1 + 2kπ

Quote:

b. .4·tan(3x - 4) .= .0

We have: .tan(3x - 4) .= .0 . . 3x - 4 .= .0 + kπ . . 3x .= .4 + kπ

Therefore: .x .= .(4/3) + (kπ/3)

Quote:

c. .sin(2x - 1) .= .

We have: .sin(2x - 1) .= . . . 2x - 1 .= .arcsin(-½)

. . which has "two" values: .(7π/6) + 2kπ .and .(11π/6) + 2kπ

2x - 1 .= .(7π/6) + 2kπ . . 2x .= .(7π/6) + 1 + 2kπ . . x .= .(7π/12) + ½ + kπ

2x - 1 .= .(11π/6) + 2kπ . . 2x .= .(11π/6) + 1 + 2kπ . . x .= .(11π/12) + ½ + kπ

• Mar 15th 2007, 03:50 PM
leviathanwave
thanks!
Anyone have any ideas on the last two? specifically the last one.
• Mar 15th 2007, 04:03 PM
mankauff
Quote:

Originally Posted by leviathanwave
Hey, I'm stuck on these 5 problems and can't seem to figure out what's what. I'm supposed to express in terms of *blah* + 2k(pi) i believe. Any help appreciated. I'll continue working on them and check back later. I really need the points for these, so I kinda need to get em all right ^_^

1. Find all solutions of the following. Express in radians.

a. 2cos^2(x-3)cos(x+1) = 0

b. 4tan(3x - 4) = 0

c. sin(2x - 1) = -1/2

2. Find the exact solutions of sin(x+2)csc(x) = 3 on the interval [0, 2(pi)]. Express in radians.

3. Find all solutions of csc(x) + cot(x) = sqrt(3). express answer in radians. Check answers. Hint: rewrite in terms of sine and cosine, clear fractions by multiplying through by the denominator, and square both sides.

Okay, this is for #3:

Things to keep aware of:
csc(x) = 1/sin(x)
cot(x) = 1/tan(x) = cos(x)/sin(x)

csc(x) + cot(x) = sqrt(3)
1/sin(x) + cos(x)/sin(x) = sqrt(3)

Multiply by sin(x)

1 + cos(x) = sqrt(3)sin(x)

Square both sides

1 + 2cos(x) + cos^2(x) = 3sin^2(x)

Now using sin^2(x) = (1 - cos^2(x)):

1 + 2cos(x) + cos^2(x) = 3(1 - cos^2(x))
1 + 2cos(x) + cos^2(x) = 3 - 3cos^2(x)
4cos^2(x) + 2cos(x) - 2 = 0
(4cos(x) - 2)(cos(x) + 1) = 0

Now solve for each:
4cos(x) - 2 = 0
4cos(x) = 2
cos(x) = 1/2
x = pi/3, 5pi/3

cos(x) + 1 = 0
cos(x) = -1
x = pi

Solutions in ascending order:
pi/3
pi
5pi/3

Hope this helps! If I can do #2, I'll post in a few minutes.
• Mar 15th 2007, 06:57 PM
leviathanwave
thanks!
Yeah number 2 is a tough one, i'm still lookin for help if anyone can :)
• Mar 16th 2007, 05:44 AM
topsquark
Quote:

Originally Posted by leviathanwave
2. Find the exact solutions of sin(x+2)csc(x) = 3 on the interval [0, 2(pi)]. Express in radians.

sin(x + 2) = sin(x)cos(2) + sin(2)cos(x)
csc(x) = 1/sin(x)

So
sin(x + 2)csc(x) = cos(2) + sin(2)cot(x) = 3

Thus
cot(x) = (3 - cos(2))/sin(2)

tan(x) = sin(2)/(3 - cos(2))

I can't think of a way to get an exact solution set for this so I will merely state: