# Trigonometric equation

#### Squish

##### New member
I'm stuck in solving for x on this one:

10sin(x)-5sin(2x)=20/pi , where 0<=x<=pi

I've got a midterm tomorrow and the answer for this one is not in the back of my book. Please help.

#### tkhunny

##### Moderator
Staff member
Have you considered the sine's double angle formula? You should memorize this important formula.

Also, you may wish to divide by 5, just to make your life a little easier. Up to you. Sometimes, it's convenient not to simplify as you go along.

#### Squish

##### New member
it'll be:
2sin(x) - 2sin(x)cos(x) = 4/pi

But I can't solve that either.

#### Dr.Peterson

##### Elite Member
The pi in the denominator seems odd. Please check that you copied correctly; does it really say $$\displaystyle 10\sin(x)-5\sin(2x)=\frac{20}{\pi}$$?

#### Squish

##### New member
The pi in the denominator seems odd. Please check that you copied correctly; does it really say $$\displaystyle 10\sin(x)-5\sin(2x)=\frac{20}{\pi}$$?
Yes, I am positively sure that is how it is written.

#### Jomo

##### Elite Member
it'll be:
2sin(x) - 2sin(x)cos(x) = 4/pi

But I can't solve that either.
Now you can divide by 2.

#### Squish

##### New member
Now you can divide by 2.
Alright, so the equation would then be
sin(x) - sin(x)cos(x)=2/pi
but still there's not much to do here; as taking sin(x) as a common factor and then trying to divide it out of the equation won't work as the 2/pi would still hold it.

#### Dr.Peterson

##### Elite Member
Well, we could replace cos(x) with sqrt(1 - sin^2(x)) and solve the resulting radical equation in sin(x); but that turns into a quartic equation that, with the irrational numbers involved, doesn't seem very workable.

So here's a new question: In your context, is it possible that you are expected to find approximate solutions using technology, rather than exact solutions by algebraic means? What else would I learn if I grabbed your book from you and looked through the context? Are any special methods needed for other problems nearby?

#### Jomo

##### Elite Member
Where did this problem come from? Are you allowed to use a graphing calculator? How about estimates?

#### Otis

##### Senior Member
… I am positively sure that [I have posted it] how it is written.
Maybe the book has a typo. #### Otis

##### Senior Member
… doesn't seem very workable …is it possible that you are expected to find approximate solutions using technology, rather than exact solutions …
That was my thought, also. There are two solutions, and (according to MVR5) the exact form of the larger solution looks like this:

arctan(1/6sqrt(6)sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)))-1/6sqrt((-6sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)))(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)-72sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)))+36sqrt(6)(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3))/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3))))),-1+1/2Pi(1/6sqrt(6)sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)))-1/6sqrt((-6sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)))(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)-72sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)))+36sqrt(6)(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3))/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3)sqrt(((27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(2/3)+12)/(Pi(27Pi+3sqrt(3)sqrt(-64+27Pi^2))^(1/3))))))^3) • Squish and topsquark

#### MarkFL

##### Super Moderator
Staff member
When I initially looked at this question, after some consideration and then looking at the HUGE solutions given by W|A, I concluded that I would use a numeric root finding technique (like the Newton-Raphson method), to approximate the two roots on the given domain. I would use a graph to generate the initial seed values. #### Squish

##### New member
Sorry for the confusion, I think I should've given a lot more context. So this part of a question on my 'webassign' assignment for calculus 2. The original question was asking me to get the average f (x) of the equation which is 20/pi by obtaining the product of (1/(b-a))(the definite integral of the above equation from b to a) where a is 0 and b is pi. The definite integration is equal to 20, so dividing it by pi would make it 20/pi. The part which I am asking about DOES expect a solution of two irrational numbers approximated to 5 decimal points. That should explain why it doesn't look like a typical highschool trigonometry question. I think I am allowed to use a graphing calculator on this assignment, but I don't own one at the moment; I am not allowed to use one in my calculus 2 exams.

#### MarkFL

##### Super Moderator
Staff member
I get:

$$\displaystyle x\approx1.23822452160548715648$$

$$\displaystyle x\approx2.8081205503453334222$$

#### Dr.Peterson

##### Elite Member
There are many online equivalents to a graphing calculator, which you can use for homework; a couple have been mentioned. A simple one is Desmos.com, on which you can graph y = 10sin(x)-5sin(2x) and y = 20/pi, then click on each intersection and see the coordinates (to only three decimal places). For more detailed and complicated answers, go to WolframAlpha.com and enter "solve 10sin(x)-5sin(2x) = 20/pi".

#### Squish

##### New member
Thank you so much