# Thread: Solving For x (angle)

1. ## Solving For x (angle)

#1 Solve: sqrt(3)(cos(x)tan(x) + cos(x) = 0, where 0 <= x <= 2pi
My work:
cos(x)[sqrt(3)tan(x) + 1] = 0

cos(x) = 0
x = pi/2, 3pi/2

sqrt(3)tan(x) + 1 = 0
x = 60

#2 Solve: cos(2x) - 3sin(x) = 2, where -pi <= x <= pi
1 - (sin(x))^2 - 3sin(x) = 2
-1 - (sin(x))^2 - 3sin(x) = 0
Unsure how to continue...

#3 Solve: 2cos(x) = 2^x
Work: Tried to graph it, but didn't really work out.

2. Originally Posted by AlphaRock

#2 Solve: cos(2x) - 3sin(x) = 2, where -pi <= x <= pi
1 - (sin(x))^2 - 3sin(x) = 2
-1 - (sin(x))^2 - 3sin(x) = 0
Unsure how to continue...

$\displaystyle \cos{2x}-3\sin{x}=2$

remember

So, lets' use the third one

$\displaystyle (1-2\sin^2{x})-3\sin{x}=2$

$\displaystyle 2\sin^2{x}+3\sin{x}+1=0$

surely you can factor a quadratic expression....

3. Originally Posted by AlphaRock
#1 Solve: sqrt(3)(cos(x)tan(x) + cos(x) = 0, where 0 <= x <= 2pi
My work:
cos(x)[sqrt(3)tan(x) + 1] = 0

cos(x) = 0
x = pi/2, 3pi/2

sqrt(3)tan(x) + 1 = 0
x = 60

#2 Solve: cos(2x) - 3sin(x) = 2, where -pi <= x <= pi
1 - (sin(x))^2 - 3sin(x) = 2
-1 - (sin(x))^2 - 3sin(x) = 0
Unsure how to continue...

#3 Solve: 2cos(x) = 2^x
Work: Tried to graph it, but didn't really work out.

Interesting. brentwoodbc asked the same exact questions for #1 and #2 in this thread: http://www.mathhelpforum.com/math-he...functions.html. VonNemo19, don't you remember?

01

4. COOL!

Looks like somebody wants the scholarship too!

(These are questions from practice exam(s)...)

5. Can someone show me clearly how to do number 3? And explain why it's a decimal? (I'm used to getting theta where it is radians or degrees (1pi/2 or 90 degrees)

6. Originally Posted by AlphaRock
Can someone show me clearly how to do number 3? And explain why it's a decimal? (I'm used to getting theta where it is radians or degrees (1pi/2 or 90 degrees)

x = 0 is an obvious solution easily seen by inspection.

The other two solutions are decimal approxiations and have been found using technology eg. graphics or CAS calculator. It's not possible to express those answers in exact form.

7. Originally Posted by yeongil
Interesting. brentwoodbc asked the same exact questions for #1 and #2 in this thread: http://www.mathhelpforum.com/math-he...functions.html. VonNemo19, don't you remember?

01

I don't have a memory. My brain is made up of virtual ram. If I shut down my life is over.

8. # 3 What a weird equation, 2cosx = 2^x. By graphing it, we have four solutions.
x1 = 0.659958
x2 = -1.377810
x3 = -4.321215
x4 = -7.851817

Below is the graph of the equation, the red curve is of 2^x and the blue one is of 2cos x. The intersection is the required answer by the proposer.