# Math Help - find solution qn

1. ## find solution qn

Given that $\frac{\pi}{10}$ is 1 of the solutions for the eqn $sin3x=cos2x$ for $0, find the 2 other solutions. Leave ur ans in terms of $\pi$.

thanks for reading.

2. Remember that $\sin{3x} = 3\sin{x} - 4\sin^3{x}$ and $\cos{2x} = 1 - 2\sin^2{x}$.

Then the equation

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

is the same as

$3\sin{x} - 4\sin^3{x} = 1 - 2\sin^2{x}$

$0 = 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1$.

This is a cubic in $\sin{x}$. So let $X = \sin{x}$ to give

$0 = 4X^3 - 2X^2 - 3X + 1$.

It's easy to see that $X = 1$ satisfies this equation.

So $\sin{x} = 1$ meaning $x = \frac{\pi}{2}$ is a solution.

You are also told that $\frac{\pi}{10}$ is a solution. This is in the first quadrant. But your domain also includes the second quadrant.

So $\pi - \frac{\pi}{10} = \frac{9\pi}{10}$ is also a solution.

So your solutions are $x = \left\{\frac{\pi}{10}, \frac{\pi}{2}, \frac{9\pi}{10}\right\}$.

3. $\sin{3x} = 3\sin{x} - 4\sin^3{x}$
what formula is that?

You are also told that $\frac{\pi}{10}$ is a solution. This is in the first quadrant. But your domain also includes the second quadrant.

So $\pi - \frac{\pi}{10} = \frac{9\pi}{10}$ is also a solution.
Dun understand this part seriously.

4. I think you know that

sin(A+B) = sinAcosB + cosAsinB

If A + B = x, then sin(2x) = 2sin(x)cos(x)

If you write sin(3x) as sin(2x+x) and expand and simplify, you will get above formula.

5. I dun get ur point either.

Prove that $sin{3x} = 3\sin{x} - 4\sin^3{x}$

$sin(2x+x)=sin(2x)cos(x)+cos(2x)sin(x)$
$=2sin(x)cos(x)cos(x)+cos(2x)sin(x)$
$=2sin(x)cos^2(x)+cos(2x)sin(x)$

6. Originally Posted by stupidguy
Dun understand this part seriously.
You can have a different method.

cos(2x) = sin(3x) = cos(π/2 - 3x) Hence the general solution will be

2x = 2nπ + π/2 - 3χ or 2x = 2nπ -π/2 + 3x. The second one gives the negative x. So leave it.

5x = 2nπ + π/2. Put n = 0, 1 and 2 and find the values of x.

7. Originally Posted by stupidguy
I dun get ur point either.

Prove that $sin{3x} = 3\sin{x} - 4\sin^3{x}$

$sin(2x+x)=sin(2x)cos(x)+cos(2x)sin(x)$
$=2sin(x)cos(x)cos(x)+cos(2x)sin(x)$
$=2sin(x)cos^2(x)+cos(2x)sin(x)$
$= 2sin(x)[1-sin^2(x)] + [1 -2sin^2(x)]sin(x)$

$= 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)$

$= 3sin(x) - 4sin^3(x)$

8. Originally Posted by sa-ri-ga-ma
You can have a different method.

cos(2x) = sin(3x) = cos(π/2 - 3x) Hence the general solution will be

2x = 2nπ + π/2 - 3χ or 2x = 2nπ -π/2 + 3x. The second one gives the negative .o leave it.

5x = 2nπ + π/2. Put n = 0, 1 and 2 and find the values of x.
I was thinking of sketching 2 graphs and finding the intersections.

The first intersection is $\dfrac{\pi}{10}$ (given), the second is 0.5 (obvious).

My real doubt is how to get the last solution which is not readable from the sketch.

According to Prove It,

$\pi - \frac{\pi}{10} = \frac{9\pi}{10}$

Why?

9. you guys actually predicted and solved the qn part 2 b4 i asked which is

show that $sin3x=3sinx-4sin^3{x}$ and hence prove that

$cos(2x)-sin(3x)\equiv 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1$

i believe that part 1 which is the original qn do not require part 2 working and is also rather simple cos it is only 2 marks, though I dunno how to do.

10. OK. Now you know that

If sin(x) = 1/2, then x can be either π/6 or (π - π/6) = 5π/6

Similarly if x = π/10 satisfies the given equation, then x = (π-π/10) = 9π/10 must also satisfy the given equation.

11. You should know that $\sin{\theta}$ is positive in the first and second quadrants, and that by symmetry, $\sin{\left(\pi - \theta\right)} = \sin{\theta}$.

In this case, $\theta = \frac{\pi}{10}$.

12. these are my "sketches".

I know that A=pi/10, how do I know B=9pi/10 assuming I sketch?

In the first place, I would not know that they are of the same y-value, hence I cannot determine their x-value like what you done.

13. You can't read from your sketch that the two graphs cross at $x = 0.9\pi = \frac{9\pi}{10}$?

Look at it this way. You have a domain of $0 < x< \pi$. This means that your solutions are in the first quadrant. So any sine values are positive.

Going back to the equation that we found using the identities I listed in my first post...

$0 = 4X^3 - 2X^2 - 3X + 1$

$0 = (X - 1)(4X^2 + 2X - 1)$.

That means $4X^2 + 2X - 1 = 0$.

This gives $X = \frac{-1 \pm \sqrt{5}}{4}$ after applying the Quadratic formula.

So $\sin{x} = \frac{-1 \pm \sqrt{5}}{4}$ since $X = \sin{x}$.

Now you need to remember from your domain of $0 < x < \pi$ that $\sin{x}$ is positive.

So we can only accept

$\sin{x} = \frac{-1 + \sqrt{5}}{4}$.

The first solution $x = \frac{\pi}{10}$ is in the first quadrant.

You also have another solution in the second quadrant. How would you find it? Think about a unit circle.

Edit: In fact, here is a picture of a unit circle, with the angle of $x = \frac{\pi}{10}$. Can you see that there is another value in the second quadrant where the sine value (the vertical lengths in red) is the same? What is that angle (measured from the positive $x$ axis)?

14. Nice sketches.

From the sketches you can clearly see that at A and B, y values are the same for both the graphs. Find the corresponding values of x.

For C, y value is different but it is same for both the graphs.

15. Notice I added inverted commas for sketches because in the exams we do not use graphic calculator.

So assuming I sketched the 2 graphs, point A and B will not be accurate (not having the same y value).

I cant read those values off my actual sketch so how do i reason it?

Page 1 of 2 12 Last