Trigonometric equations question

[spoc21]

Homework Statement

Solve for x for $$0\leq x$$$$\leq 2 \pi$$ in the following equation:

sin x tan x + tan x - 2sinx + cos x =0

Homework Equations

N/A

The Attempt at a Solution

I tried making $$tan x = \frac{cos x}{sin x}$$, but did not have any luck, and am very confused on how to solve the equation.

I would really appreciate any help/suggestions.

Thanks

 

[phyzguy]

Well, actually $$tan(x) = \frac{sin(x)}{cos(x)}$$. Try this, put everything over a common denominator, and simplify

 

[spoc21]

Yeah, sorry, my mistake
I did actually try $$tan x = \frac{cos x}{sin x}$$, and tried to solve for a common denominator, cos (x), but am still very confused.

 

[Mark44]

This seems to be a tough one. If you replace the tan expressions you get
$$\frac{sin^2 x}{cos x} + \frac{sin x}{cos x} -2sin x + cos x = 0$$

Multiplying by cos x gives you
$$sin^2 x + sin x -2sin x cos x + cos^2 x = 0$$

Which becomes
$$1 + sin x -2sin x cos x = 0$$

I haven't been able to get to anything reasonable beyond this point. Can you confirm that we are working the right problem?

 

[phyzguy]

If you keep going, you can write this as: 1+sin(x) = sin(2x). This has one solution that can be obtained by inspection, namely x = 3*pi/2, where sin(2x)=0, and sin(x)=-1. By graphing it out you can see that there is a second solution with a numerical value of x~=3.49, but I don't know how to get an analytic value for this solution.

 

[spoc21]

This seems to be a tough one. If you replace the tan expressions you get
$$\frac{sin^2 x}{cos x} + \frac{sin x}{cos x} -2sin x + cos x = 0$$

Multiplying by cos x gives you
$$sin^2 x + sin x -2sin x cos x + cos^2 x = 0$$

Which becomes
$$1 + sin x -2sin x cos x = 0$$

I haven't been able to get to anything reasonable beyond this point. Can you confirm that we are working the right problem?

Yes, this is precisely where I'm stuck. I'm not sure on how to proceed after this step.

 

[Mark44]

If you keep going, you can write this as: 1+sin(x) = sin(2x).

Yeah, I got that, too, but thought I should be able to get an equation that could be solved by means other than inspection.

This has one solution that can be obtained by inspection, namely x = 3*pi/2, where sin(2x)=0, and sin(x)=-1. By graphing it out you can see that there is a second solution with a numerical value of x~=3.49, but I don't know how to get an analytic value for this solution.

 

[abclemons]

If you keep going, you can write this as: 1+sin(x) = sin(2x). This has one solution that can be obtained by inspection, namely x = 3*pi/2, where sin(2x)=0, and sin(x)=-1. By graphing it out you can see that there is a second solution with a numerical value of x~=3.49, but I don't know how to get an analytic value for this solution.

I think at this point the "analytical" thing to do would be to set each side equivalent to 0 and solve. So that:

1+sin(x)=0
sin(x)=-1
x=3$$\pi$$/2

2sin(x)cos(x)=0
sin(x)=0
x=$$\pi$$ and 2$$\pi$$

2sin(x)cos(x)=0
cos(x)=0
x=3$$\pi$$/2

At this point, I think it becomes a plug-and-chug with 3$$\pi$$/2 being the only viable option.

 

[Mark44]

Sorry, that doesn't make any sense to me. a = b does not necessarily imply that a = 0 or that b = 0.

 

[phyzguy]

If you really want an analytic solution, you can square both sides to obtain the following:
$$1+2sin(x)+sin^2(x)=sin^2(2x)=1-cos^2(2x)=1-(1-2sin^2(x))^2=4sin^2(x)-4sin^4(x)$$
collecting terms and calling s = sin(x) gives:
$$4s^4-3s^2+2s+1=0$$
This can be solved using known techniques (I actually used Mathematica), with the only real solutions being s=-1 (which is the 3*pi/2 solution for x), and:
s=1/6 (2 - 1/(28 - 3 Sqrt[87])^(1/3) - (28 - 3 Sqrt[87])^(1/3))
So for x, we have:
x = 3*pi/2
x = arcsin(1/6 (2 - 1/(28 - 3 Sqrt[87])^(1/3) - (28 - 3 Sqrt[87])^(1/3)))

Since we squared both sides, we introduced some extra solutions, so we have to sort through the multiple values of the arcsin to find the correct one, and realize that only the one in quadrant 3 gives an actual solution. This is the x~3.49 solution from the graphical analysis.