1. ## general solution?

determine the general solution:

2(sinxcosx - 1/2) =0

2} Write down the equation of q, if q is the result of p shifted 3 units to the right.

2. Originally Posted by Viky
determine the general solution:

2(sinxcosx - 1/2) =0
Note that when you distribute the 2, you have

$2\sin x\cos x-1=0\implies \sin(2x)=1$

Thus, $2x=\sin^{-1}(1)\implies 2x=\frac{(4k-3)\pi}{2};\,k\in\mathbb{Z}$

So it follows that $x=\frac{(4k-3)\pi}{4};\,k\in\mathbb{Z}$

Does this make sense?

2} Write down the equation of q, if q is the result of p shifted 3 units to the right.
When you shift a function $f(x)$ $n$ units to the right, you generate a function $g(x)=f(x-n)$. Use this idea to answer this part of the question.

3. i didn't get the first part

4. ## General solution of trig equation

Hello Viky
Originally Posted by Viky
i didn't get the first part
$2(\sin x\cos x -\tfrac12)=0$

$\Rightarrow 2\sin x \cos x -1 =0$

$\Rightarrow \sin 2x = 1$, using the identity $\sin 2x = 2 \sin x \cos x$

Since $\sin \frac{\pi}{2} = 1$, one possible solution is $2x = \frac{\pi}{2}$; i.e. $x = \frac{\pi}{4}$. But how do we find the general solution? Like this:

The positive values of $\theta$ that make $\sin\theta = 1$ are $\frac{\pi}{2},\frac{\pi}{2}+ 2\pi,\frac{\pi}{2}+ 4\pi, \frac{\pi}{2}+ 6\pi$, and so on.

Or we can go the other way and find negative values of $\theta$ by taking away multiples of $2\pi$. So $\sin\theta = 1$ is also satisfied by $\theta = \frac{\pi}{2}- 2\pi, \frac{\pi}{2}-4\pi$, and so on.

Combining all these together we can say that the general solution of $\sin\theta = 1$ is:

$\theta = \frac{\pi}{2}+ 2k\pi, k = 0, \pm1, \pm2, \pm3, ...$

or simply $\theta = \frac{\pi}{2}+ 2k\pi, k \in \mathbb{Z}$, where $\mathbb{Z} = \{\text{integers}\}$

Now replace $\theta$ by $2x$, and we get

$\sin 2x = 1$

$\Rightarrow 2x = \frac{\pi}{2}+ 2k\pi, k \in \mathbb{Z}$

Divide by 2:

$\Rightarrow x = \frac{\pi}{4}+ k\pi, k \in \mathbb{Z}$

This is equivalent to the answer that Chris L T521 gave you, but the values of $k$ start in a different place.