Solve the trigonometric equation sin(2theta-pi/6) = cos(2theta)

[iixfaith]

View attachment 5897

i need help with question 4a.
Thank to anyone who helped.

 

[MarkFL]

Re: need help with this Trigo question 4a

We are given to solve:

\(\displaystyle \sin\left(2\theta-\frac{\pi}{6}\right)=\cos(2\theta)\) for $0\le\theta\le2\pi$

I would begin by looking at the angle difference identity for sine:

\(\displaystyle \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\)

Applying that identity, what do you get for the equation?

 

[iixfaith]

Re: need help with this Trigo question 4a

We are given to solve:

\(\displaystyle \sin\left(2\theta-\frac{\pi}{6}\right)=\cos(2\theta)\) for $0\le\theta\le2\pi$

I would begin by looking at the angle difference identity for sine:

\(\displaystyle \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\)

Applying that identity, what do you get for the equation?

i would be getting this:

View attachment 5898

 

[MarkFL]

Please crop and attach your images inline, as those free image hosting sites don't work well at all...I see only blank space when you link to them, and so I have to edit your post to load the image URL into my browser, save the image to my hard drive, resize it, crop it, and then attach it and save the edited post so that everyone will be able to see the images.

Okay, can you simplify, as you should know the values of $\sin\left(\frac{\pi}{6}\right)$ and $\cos\left(\frac{\pi}{6}\right)$. :)

 

[iixfaith]

Please crop and attach your images inline, as those free image hosting sites don't work well at all...I see only blank space when you link to them, and so I have to edit your post to load the image URL into my browser, save the image to my hard drive, resize it, crop it, and then attach it and save the edited post so that everyone will be able to see the images.

Okay, can you simplify, as you should know the values of $\sin\left(\frac{\pi}{6}\right)$ and $\cos\left(\frac{\pi}{6}\right)$. :)

$\sin\left({2\theta}\right)\frac{\sqrt{3}}{2}-\cos\left({2\theta}\right)\frac{1}{2}$

 

[MarkFL]

$\sin\left({2\theta}\right)\frac{\sqrt{3}}{2}-\cos\left({2\theta}\right)\frac{1}{2}$

Okay, good! (Yes)

So, if we multiply through by $2$, the equation becomes:

\(\displaystyle \sqrt{3}\sin(2\theta)-\cos(2\theta)=2\cos(2\theta)\)

What do you get now if you add $\cos(2\theta)$ to both sides?

 

[iixfaith]

Okay, good! (Yes)

So, if we multiply through by $2$, the equation becomes:

\(\displaystyle \sqrt{3}\sin(2\theta)-\cos(2\theta)=2\cos(2\theta)\)

What do you get now if you add $\cos(2\theta)$ to both sides?

is it ?

$\sqrt{3}\sin\left({2\theta}\right) = 2\cos\left({2\theta}\right)/+\cos\left({2\theta}\right)$

 

[MarkFL]

is it ?

$\sqrt{3}\sin\left({2\theta}\right) = 2\cos\left({2\theta}\right)/+\cos\left({2\theta}\right)$

You can combine the 2 terms on the right so that you have:

\(\displaystyle \sqrt{3}\sin(2\theta)=3\cos(2\theta)\)

we have lots of choices at this point on how to proceed...my choice would be to divide through by:

\(\displaystyle \sqrt{3}\cos(2\theta)\ne0\) to obtain:

\(\displaystyle \tan(2\theta)=\sqrt{3}\)

Now, if given:

\(\displaystyle \tan(u)=\sqrt{3}\)

and knowing that $\tan(u)=\tan(u+k\pi)$ where $k\in\mathbb{Z}$

then we can state:

\(\displaystyle u=\frac{\pi}{3}+k\pi=\frac{\pi}{3}(3k+1)\)

And so applying this to our equation, we may state:

\(\displaystyle 2\theta=\frac{\pi}{3}(3k+1)\)

And upon dividing through by $2$, we obtain:

\(\displaystyle \theta=\frac{\pi}{6}(3k+1)\)

Now, you just need to find the values for $k$ which put $\theta$ in the given interval:

\(\displaystyle 0\le\theta\le2\pi\)

Substitute for $\theta$:

\(\displaystyle 0\le\frac{\pi}{6}(3k+1)\le2\pi\)

Multiply through by \(\displaystyle \frac{6}{\pi}\)

\(\displaystyle 0\le3k+1\le12\)

\(\displaystyle -1\le3k\le11\)

\(\displaystyle -\frac{1}{3}\le k\le\frac{11}{3}\)

Hence:

\(\displaystyle k\in\{0,1,2,3\}\)

Now with:

\(\displaystyle \theta(k)=\frac{\pi}{6}(3k+1)\)

We may give the 4 solutions to the equation as:

\(\displaystyle \theta(0)=\frac{\pi}{6}(3(0)+1)=\frac{\pi}{6}\)

\(\displaystyle \theta(1)=\frac{\pi}{6}(3(1)+1)=\frac{2\pi}{3}\)

\(\displaystyle \theta(2)=\frac{\pi}{6}(3(2)+1)=\frac{7\pi}{6}\)

\(\displaystyle \theta(3)=\frac{\pi}{6}(3(3)+1)=\frac{5\pi}{3}\)

Does this make sense?

 

[iixfaith]

You can combine the 2 terms on the right so that you have:

\(\displaystyle \sqrt{3}\sin(2\theta)=3\cos(2\theta)\)

we have lots of choices at this point on how to proceed...my choice would be to divide through by:

\(\displaystyle \sqrt{3}\cos(2\theta)\ne0\) to obtain:

\(\displaystyle \tan(2\theta)=\sqrt{3}\)

Now, if given:

\(\displaystyle \tan(u)=\sqrt{3}\)

and knowing that $\tan(u)=\tan(u+k\pi)$ where $k\in\mathbb{Z}$

then we can state:

\(\displaystyle u=\frac{\pi}{3}+k\pi=\frac{\pi}{3}(3k+1)\)

And so applying this to our equation, we may state:

\(\displaystyle 2\theta=\frac{\pi}{3}(3k+1)\)

And upon dividing through by $2$, we obtain:

\(\displaystyle \theta=\frac{\pi}{6}(3k+1)\)

Now, you just need to find the values for $k$ which put $\theta$ in the given interval:

\(\displaystyle 0\le\theta\le2\pi\)

Substitute for $\theta$:

\(\displaystyle 0\le\frac{\pi}{6}(3k+1)\le2\pi\)

Multiply through by \(\displaystyle \frac{6}{\pi}\)

\(\displaystyle 0\le3k+1\le12\)

\(\displaystyle -1\le3k\le11\)

\(\displaystyle -\frac{1}{3}\le k\le\frac{11}{3}\)

Hence:

\(\displaystyle k\in\{0,1,2,3\}\)

Now with:

\(\displaystyle \theta(k)=\frac{\pi}{6}(3k+1)\)

We may give the 4 solutions to the equation as:

\(\displaystyle \theta(0)=\frac{\pi}{6}(3(0)+1)=\frac{\pi}{6}\)

\(\displaystyle \theta(1)=\frac{\pi}{6}(3(1)+1)=\frac{2\pi}{3}\)

\(\displaystyle \theta(2)=\frac{\pi}{6}(3(2)+1)=\frac{7\pi}{6}\)

\(\displaystyle \theta(3)=\frac{\pi}{6}(3(3)+1)=\frac{5\pi}{3}\)

Does this make sense?

Thanks ! i have been stuck at this question for at least 2 days finally able to understand and solve thanks to you

 

[soroban]

$$\text{4(a). Solve: }\:\sin(2\theta - \tfrac{\pi}{6}) \:=\:\cos2\theta\; \text{ for }0 \le \theta \le 2\pi.$$

$$\begin{array}{ccc} \sin(2\theta - \frac{\pi}{6}) \;=\; \cos(2\theta) \\ \\ \sin(2\theta) \cos(\frac{\pi}{6}) - \sin(\frac{\pi}{6})\cos(2\theta) \;=\;\cos(2\theta) \\ \\ \frac{\sqrt{3}}{2}\sin(2\theta) - \frac{1}{2}\cos(2\theta) \;=\;\cos(2\theta) \\ \\ \frac{\sqrt{3}}{2}\sin(2\theta) \;=\;\frac{3}{2}\cos(2\theta) \\ \\ \dfrac{\sin(2\theta)}{\cos(2\theta)} \;=\;\dfrac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} \\ \\ \tan(2\theta) \;=\;\sqrt{3} \\ \\ 2\theta \;=\;\dfrac{\pi}{3},\;\dfrac{4\pi}{3},\;\dfrac{7\pi}{3},\;\dfrac{10\pi}{3} \\ \\ \theta \;=\;\dfrac{\pi}{6},\;\dfrac{2\pi}{3},\;\dfrac{7\pi}{6},\;\dfrac{5\pi}{3} \end{array}$$

 

[MarkFL]

soroban, going from your 3rd to 4th lines of work you have an error in which the RHS of the 4th line should be:

\(\displaystyle \frac{3}{2}\cos(2\theta)\)

You have essentially solved:

\(\displaystyle \sin\left(2\theta-\frac{\pi}{6}\right)=0\)