Trig equation: cos(3π/2−2x)+sin(π−3x)=sin(3x)−cos(x)

You can see question in photo below

Please tell me identities and/or other stuff to solve this.

Tuharramah said:

You can see question in photo below

Please tell me identities and/or other stuff to solve this.

Click to expand...

Okay, just so we'll all on the same page, this is what I think your image says the problem is. If it's incorrect, please reply with any necessary corrections:

\(\displaystyle cos \left( \dfrac{3\pi}{2}-2x \right) + sin \left( \pi - 3x \right) = sin(3x) - cos(x)\)

I'll assume that you've read the Read Before Postinghttps://www.freemathhelp.com/forum/threads/41537-Read-Before-Posting!! thread, and also assume that because you've shown no work of your own, you have none to share with us and need help at the very beginning. I'd begin by revisiting your class notes and/or textbook to find any identities you've been taught in the class so far. In particular, I'd look for the sum and difference identities. They should look something like: \(\displaystyle sin(\alpha + \beta) = \text{...}\). If, for whatever reason, you can't find them in your notes or textbook, you can try this webpagehttps://brownmath.com/twt/sumdiff.htm for a refresher.

Once you've reviewed these identities, go ahead and give the problem your best effort. Then please include any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.

By studying your suggestion

I got x as 30 deg.
Unfortunately i couldn't upload my solution
So is it correct answer?

Tuharramah said:

I got x as 30 deg.
Unfortunately i couldn't upload my solution
So is it correct answer?

Click to expand...

If you put x = 30°, do you satisfy the given equation:

No

< link to objectionable site removed >

Here is my solution
As I said I couldn't upload it here

Your solution of \(\displaystyle x = 30^{\circ}=\dfrac{\pi}{6}\) is a valid solution...

\(\displaystyle cos \left( \dfrac{3\pi}{2} - \dfrac{2 \pi}{6} \right) + sin \left( \pi - \dfrac{3 \pi}{6} \right) = sin\left( \dfrac{3 \pi}{6} \right) - cos\left( \dfrac{\pi}{6} \right)\)

\(\displaystyle cos \left( \dfrac{7 \pi}{6} \right) + sin \left( \dfrac{\pi}{2} \right) = sin\left( \dfrac{3 \pi}{6} \right) - cos\left( \dfrac{\pi}{6} \right)\)

\(\displaystyle -\dfrac{\sqrt{3}}{2} + 1 = 1 - \dfrac{\sqrt{3}}{2}\)

\(\displaystyle 0.134 \approx 0.134\)

...but it's not the only one. I'm assuming you're only interested in solutions for \(\displaystyle 0 \le x \le 2\pi\) or \(\displaystyle 0^{\circ} \le x \le 360^{\circ}\). There's another solution to the equation \(\displaystyle sin(x)=\dfrac{1}{2}\) on that interval. Plus, there's two other solutions this equation won't find you.

Your work is fine up until you get to this point:

\(\displaystyle -2sin(x)cos(x) + cos(x) = 0\)

From here, what happens if you multiply by -1 and factor?

\(\displaystyle cos(x) \cdot (2sin(x) - 1) = 0\)

Where does that lead you?

So it will be:
-2sin (x)×cos (x)+cos (x)=0

2sin (x)×cos (x)-cos (x)=0

Cos (x) × (2sin (x) -1)=0

X₁=90
X₂=30
..
So when I divided by cos x ,I killed one solution

Tuharramah said:

So it will be:
-2sin (x)×cos (x)+cos (x)=0

2sin (x)×cos (x)-cos (x)=0

Cos (x) × (2sin (x) -1)=0

X₁=90
X₂=30
..
So when I divided by cos x ,I killed one solution

Click to expand...

Was your domain restricted like: 0 ≤ x ≤ π/2

Let's suppose it is.
Because no information about domain is given

I don't know about your class and teacher specifically, but my experience has always been that if the problem doesn't specify a domain, they want all the solutions everywhere. If you've found the four solutions on [0, 2pi], use the fact that sine and cosine have period 2pi (360 degrees) to write four "families" of solutions.