Plse help me find the general solution to this problem

[val1]

I am trying to find the general solution in radians to the following equation:

tanx/2 = sinxcosx

This is how far I've got.

2sinxcosx - tanx = 0

sinx(2cosx) - tanx = 0

To make sinx = 0, x= 0, pi or 2pi for values of x between 0 and 2pi.

To make 2cosx = 0, x = pi/2, 3pi/2 for values of x between 0 and 2pi.

But i don't know how to deal with the tanx in the expression.

I've tried to make tanx = 0 , by making x =0, or any value of n.pi (where n is an integer) but this doesn't satisfy the equation when n.pi is substituted elsewhere in the original equation.

Please can you say where am I going wrong?

Thanks

 

[pka]

Because \(\displaystyle \tan (x) = \frac{{\sin (x)}}{{\cos (x)}}\) any x for which sin(x)=0 is a solution: \(\displaystyle x = k\pi\) where k is an integer.

Now lets look at the case \(\displaystyle \sin (x) \ne 0\) that means \(\displaystyle \frac{1}{{2\cos (x)}} = \cos (x)\) or \(\displaystyle \cos ^2 (x) = \frac{1}{2}\).

That happens if \(\displaystyle x = (\pi /4) + k\pi\) or \(\displaystyle x = ( - \pi /4) + k\pi\) where k is an integer.

Please note that in your work, cos(x) CANNOT be 0.

 

[soroban]

Hello, val1!

If that is really \(\displaystyle \frac{\tan x}{2}\) . . . and not \(\displaystyle \tan(\frac{x}{2})\) . . . it's easier . . .

\(\displaystyle \frac{\tan x}{2}\:=\:\sin x\cdot\cos x\)

We have: .\(\displaystyle \frac{\sin x}{2\cdot \cos x}\:=\:\sin x\cdot\cos x\;\;\Rightarrow\;\;\sin x\:=\:2\cdot\sin x\cdot\cos^2x\)

. . \(\displaystyle \sin x\:=\:2\cdot\sin x\cdot(1\,-\,\sin^2x)\;\;\Rightarrow\;\;\sin x\:=\:2\cdot\sin x - 2\cdot\sin^3x\)

. . \(\displaystyle 2\cdot\sin^3x - \sin x\:=\:0\;\;\Rightarrow\;\;\sin x\cdot(2\cdot\sin^2x\,-\,1)\:=\:0\)


And we have two equations to solve:

. . \(\displaystyle \sin x\,=\,0\;\;\Rightarrow\;\;x\,=\,n\pi\)

. . \(\displaystyle 2\cdot\sin^2x - 1\,=\,0\;\;\Rightarrow\;\;\sin x\,=\,\pm\frac{1}{\sqrt{2}}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4}\,+\,(\frac{\pi}{2})n\)

 

[stapel]

...find the general solution in radians to the following equation:
tanx/2 = sinxcosx
This is how far I've got.
2sinxcosx - tanx = 0
sinx(2cosx) - tanx = 0
...Please can you say where am I going wrong?

You made a good start -- and thank you for showing your work.

To answer your question, I think you went astray when you attempted to solve the equation by setting every term equal to zero. Granted, (0)(0) - 0 does indeed equal 0, but you can't know (at least, not in advance) that there is any one x for which all of sin(x), cos(x), and tan(x) will be zero (and, in fact, there isn't such an x).

In general, the method for solving this sort of exercise is to convert one side to something that factors, and then solve the factors. It's like when you had "x² - 2x = 0", and you solved by converting to "x(x - 2) = 0", and then solving the factors to get x = 0 and x = 2. Do the same here:

. . . . .2sin(x)cos(x) - tan(x) = 0

Note that tan(x) = [sin(x)]/[cos(x)]. Than:

. . . . .2sin(x)cos(x) - sin(x)/cos(x) = 0

Note: You can't have cos(x) = 0 (or you'd be dividing by zero).

Convert to a common denominator:

. . . . .[2sin(x)cos(x)] / cos(x) - sin(x) / cos(x) = 0

Combine:

. . . . .[2sin(x)cos(x) - sin(x)] / cos(x) = 0

A fraction is zero when the numerator is zero, so you're really solving:

. . . . .2sin(x)cos(x) - sin(x) = 0

Now factor and solve in the usual way.

Eliz.