Given that $\displaystyle \frac{\pi}{10}$ is 1 of the solutions for the eqn $\displaystyle sin3x=cos2x $ for $\displaystyle 0<x<\pi$, find the 2 other solutions. Leave ur ans in terms of $\displaystyle \pi$.
thanks for reading.
Remember that $\displaystyle \sin{3x} = 3\sin{x} - 4\sin^3{x}$ and $\displaystyle \cos{2x} = 1 - 2\sin^2{x}$.
Then the equation
$\displaystyle \sin{3x} = \cos{2x}$
is the same as
$\displaystyle 3\sin{x} - 4\sin^3{x} = 1 - 2\sin^2{x}$
$\displaystyle 0 = 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1$.
This is a cubic in $\displaystyle \sin{x}$. So let $\displaystyle X = \sin{x}$ to give
$\displaystyle 0 = 4X^3 - 2X^2 - 3X + 1$.
It's easy to see that $\displaystyle X = 1$ satisfies this equation.
So $\displaystyle \sin{x} = 1$ meaning $\displaystyle x = \frac{\pi}{2}$ is a solution.
You are also told that $\displaystyle \frac{\pi}{10}$ is a solution. This is in the first quadrant. But your domain also includes the second quadrant.
So $\displaystyle \pi - \frac{\pi}{10} = \frac{9\pi}{10}$ is also a solution.
So your solutions are $\displaystyle x = \left\{\frac{\pi}{10}, \frac{\pi}{2}, \frac{9\pi}{10}\right\}$.
what formula is that?$\displaystyle \sin{3x} = 3\sin{x} - 4\sin^3{x}$
Dun understand this part seriously.
You are also told that $\displaystyle \frac{\pi}{10}$ is a solution. This is in the first quadrant. But your domain also includes the second quadrant.
So $\displaystyle \pi - \frac{\pi}{10} = \frac{9\pi}{10}$ is also a solution.
I was thinking of sketching 2 graphs and finding the intersections.
The first intersection is $\displaystyle \dfrac{\pi}{10}$ (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,
$\displaystyle \pi - \frac{\pi}{10} = \frac{9\pi}{10}$
Why?
you guys actually predicted and solved the qn part 2 b4 i asked which is
show that $\displaystyle sin3x=3sinx-4sin^3{x}$ and hence prove that
$\displaystyle cos(2x)-sin(3x)\equiv 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1$
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.
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.
You should know that $\displaystyle \sin{\theta}$ is positive in the first and second quadrants, and that by symmetry, $\displaystyle \sin{\left(\pi - \theta\right)} = \sin{\theta}$.
In this case, $\displaystyle \theta = \frac{\pi}{10}$.
You can't read from your sketch that the two graphs cross at $\displaystyle x = 0.9\pi = \frac{9\pi}{10}$?
Look at it this way. You have a domain of $\displaystyle 0 < x< \pi$. 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...
$\displaystyle 0 = 4X^3 - 2X^2 - 3X + 1$
$\displaystyle 0 = (X - 1)(4X^2 + 2X - 1)$.
That means $\displaystyle 4X^2 + 2X - 1 = 0$.
This gives $\displaystyle X = \frac{-1 \pm \sqrt{5}}{4}$ after applying the Quadratic formula.
So $\displaystyle \sin{x} = \frac{-1 \pm \sqrt{5}}{4}$ since $\displaystyle X = \sin{x}$.
Now you need to remember from your domain of $\displaystyle 0 < x < \pi$ that $\displaystyle \sin{x}$ is positive.
So we can only accept
$\displaystyle \sin{x} = \frac{-1 + \sqrt{5}}{4}$.
The first solution $\displaystyle x = \frac{\pi}{10}$ 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 $\displaystyle x = \frac{\pi}{10}$. 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 $\displaystyle x$ axis)?
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?