1. ## 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.
20170619_191130.jpg

2. Originally Posted by Tuharramah
You can see question in photo below

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

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

I'll assume that you've read the Read Before 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: $sin(\alpha + \beta) = \text{...}$. If, for whatever reason, you can't find them in your notes or textbook, you can try this webpage 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.

3. Originally Posted by Tuharramah
You can see question in photo below

Please tell me identities and/or other stuff to solve this.
20170619_191130.jpg
cos(3π/22x)+sin(π3x)=sin(3x)cos(x)

Useful identities for this problem:

cos(π/2 - Θ) = ?

cos(π - Θ) = ?

sin(π - Θ) = ?

sin(3Θ) = ? [as a function of sin(Θ) and cos(Θ)]

cos(2Θ) = ? [as a function of sin(Θ) and cos(Θ)]

4. ## By studying your suggestion

I got x as 30 deg.
Unfortunately i couldn't upload my solution

5. Originally Posted by Tuharramah
I got x as 30 deg.
Unfortunately i couldn't upload my solution
If you put x = 30°, do you satisfy the given equation:

6. ## No :(

< link to objectionable site removed >

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

7. Your solution of $x = 30^{\circ}=\dfrac{\pi}{6}$ is a valid solution...

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

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

$-\dfrac{\sqrt{3}}{2} + 1 = 1 - \dfrac{\sqrt{3}}{2}$

$0.134 \approx 0.134$

...but it's not the only one. I'm assuming you're only interested in solutions for $0 \le x \le 2\pi$ or $0^{\circ} \le x \le 360^{\circ}$. There's another solution to the equation $sin(x)=\dfrac{1}{2}$ 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:

$-2sin(x)cos(x) + cos(x) = 0$

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

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

8. 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

9. Originally Posted by Tuharramah
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
Was your domain restricted like: 0 ≤ x ≤ π/2

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

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