Question

Hi, i'm struggling trying to figure out how to solve this math equation:
4sinxtanx - cosxcotx = 0.

The final answer was given on the back of the sheet.
I should get x = 32 degrees and 212 degrees because of CAST. Anyone able to help with this?

Answer 1

Rewrite everything in terms of sines and cosines.

\[\large \tan x=\frac{\sin x}{\cos x}\]
\[\large \cot x=\frac{\cos x}{\sin x}\]

If we plug these in for our tangent and cotangent, and cancel some stuff out, what equation does it leave us with?

Answer 2

4sinx(sinx/cosx) - cosx(cosx/sinx) = 0
From there It became:

(4sin^2x/cosx) - (cos^2x/sinx) = 0.

Now I tried to find a common denominator, turning the equation into:

4sin^3x - cos^3x/ cosxsinx

or another way to write that would be:

4sin^3x/cosxsinx - cos^3x/cosxsins = 0

Answer 3

Multiply both sides by the common denominator, giving us,\[\large 4\sin^3x-\cos^3x=0\]Difference of cubes.. hmm yah this looks tricky :o hmmmm

Answer 4

From my understanding I cannot just cancel out the bottom from the top because of the subtraction sign in the middle. And if I did cancel stuff out, it would return me to the previous step.

Answer 5

Let's look at it from this step,\[\large \frac{4\sin^2x}{\cos x}-\frac{\cos^2x}{\sin x}=0\]Multiplying both sides by sin x gives us,\[\large \sin x\left(\frac{4\sin^2x}{\cos x}-\frac{\cos^2x}{\sin x}\right)=0 \qquad \rightarrow \qquad \frac{4\sin^3x}{\cos x}-\cos^2x=0\]Multiplying both sides by cos x gives us,\[\large \cos x\left(\frac{4\sin^3x}{\cos x}-\cos^2x\right)=0 \qquad \rightarrow\qquad 4\sin^3x-\cos^3x=0\]

Because of the subtraction? D: I'm not sure what you mean.

Answer 6

4sin^3x - cos^3x should both be over cosxsinx because it is the common denominator.

Answer 7

Ok if we're going to get a common denominator, then from that step,\[\large \frac{4\sin^3x-\cos^3x}{\sin x \cos x}=0\]
Multiplying both sides by sin x cos x gives us,\[\large \cancel{\sin x \cos x}\left(\frac{4\sin^3x-\cos^3x}{\cancel{\sin x \cos x}}\right)=(0)\sin x \cos x\]

Answer 8

And then solve for x: cot(x)=4^(1/3)

Answer 9

oops, first suppose sin(x)=0, then cos(x)=1, and LHS =/= RHS => sin(x) =/= 0

Answer 10

cos(x)=-1 or 1, either of them leads to a discrepancy between LHS/RHS

Answer 11

Alright I see what you mean now.

So what I think I could do now then is divide both \[4\sin^3x \] and \[\cos^3x\]

by cos^3x

Turning what I have into:

(4sin^3x/cos^3x) - (cos^3x/cos^3x)

Which I do believe can become:

4tan^3x - 1 = 0.

I bring the 1 and for over:

tan^3x = 1/4

Then I do:

\[\sqrt[3]{?}\]

Which leaves me with:

Tanx = \[\sqrt[3]{1/4}\]

becoming finally: 0.6299

So I just take the tan inverse of 0.6299 and round to the nearest degree and I get 32.

I know I just figured out the rest of it. But thanks for the help given!

Answer 12

You can either divide by cos(x)^3 or sin(x)^3, but make sure the you are not falling into the trap of dividing by zero