# How to solve simple trigonometry equation.

So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)

• $\sin 3x \cdot \cos 3x = \sin 2x$

• $2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 x) - \cos x = 0$

• $3 \sin^2 x - 4 \sin x \cdot \cos x + 5 \cos^2 x = 2$

• $\sin^2 x - \sin^4 x + \cos^4 x = 1$

In our students book they're poorly explained with 2 pages, I tried to find a solution on the web, but still couldn't find similar examples. All we got from our teacher was paper with few formulas and we basically have no idea when to use them. I would show what I've tried, but the problem is that I have no idea to even start solving such equations.

Are your formulae things like $\sin^2 x + \cos^2 x = 1$?

If so then you need to spot where you can apply them.

For example $$\sin^2 x - \sin^4 x + \cos^4 x = 1$$ $$\sin^2 x (1- \sin^2 x) + \cos^4 x = 1$$ $$\sin^2 x \cos^2 x + \cos^4 x = 1$$ $$(\sin^2 x + \cos^2 x) \cos^2 x = 1$$ $$\cos^2 x = 1$$ $$\cos x = \pm 1$$

• That's great! Thanks! How about the first one? I couldn't find any similar formulas. Also the third one at first look seemd to be (a + b)^2 formula, but it isn't. What I do then? – Deepsy Feb 26 '14 at 0:18
• For the first you may have formulae for $\sin(x+y)$ etc. – Henry Feb 26 '14 at 0:33

For the third one $$3\sin^2x-4\sin x\cos x+5\cos^2x=2\\ \Rightarrow\sin^2x-4\sin x\cos x+4\cos^2x=2-2\sin^2x-\cos^2x\\ \Rightarrow(\sin x-2\cos x)^2=2(1-\sin^2x)-\cos^2x\\ \Rightarrow(\sin x-2\cos x)^2=2\cos^2x-\cos^2x\\ \Rightarrow(\sin x-2\cos x)^2=cos^2x\\ \Rightarrow(\sin x-2\cos x)^2-\cos^2x=0\\ \Rightarrow(\sin x-2\cos x+\cos x)(\sin x-2\cos x-\cos x)=0\\ \Rightarrow(\sin x-\cos x)(\sin x-3\cos x)=0$$ Then $$\sin x-\cos x=0\Rightarrow\tan x=1$$ and $$\sin x-3\cos x=0\Rightarrow\tan x=3$$

For the first one I don't know how far your book is covering. I have used the following two rules for that $$\sin 2x=2\sin x\cos x$$ and $$\sin 3x=3\sin x-4\sin^3x$$ So it goes like this $$\sin 3x .\cos 3x=\sin2x\\ \Rightarrow 2\sin 3x .\cos 3x=2\sin2x\\ \Rightarrow\sin6x=2\sin2x\\ \Rightarrow\sin3(2x)=2\sin2x\\ \Rightarrow 3\sin2x-4\sin^32x=2\sin2x\\ \Rightarrow 3\sin2x-4\sin^32x-2\sin2x=0\\ \Rightarrow \sin2x-4\sin^32x=0\\ \Rightarrow \sin2x(1-4\sin^22x)=0\\$$ Then $$\sin 2x=0$$ and $$\sin2x=\pm \frac{1}{2}$$ Hope it helps.

For the second:

$$2\left(1+\sin^6 x+\cos^6 x\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x= 0 \\ 2\left(1+\left( \sin^2 x + \cos^2 x \right) \left(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x\right)\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x = 0 \\ 2\left(1+\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x= 0 \\ 2+2\sin^4 x - 2\sin^2 x \cos^2 x + 2\cos^4 x-3\sin^4 x -3\cos^4 x - \cos x= 0 \\ 2-\sin^4 x - 2\sin^2 x \cos^2 x - \cos^4 x - \cos x= 0 \\ 2-\left(\sin^4 x + 2\sin^2 x \cos^2 x + \cos^4 x \right)- \cos x= 0 \\ 2-\left(\sin^2 x + \cos^2 x \right)^2- \cos x= 0 \\ 2-1- \cos x= 0 \\ \cos x=1 \\$$