1. ## trig equations

hi guys , i am a high school student , and i have a trigonometry test tomorrow . i would appreciate every help from you.
for now i can't solve these trigonometry equations.
they are mashing up my mind

1) sin2x + sinx * cosx = 1

2)1+cosx + cos(x/2) =0

3) 1 + cos(π +x ) + cos(π/2 + x/2) =0

2. ## Re: trig equations

So you just want someone else to do them for you?

3. ## Re: trig equations

Originally Posted by HallsofIvy
So you just want someone else to do them for you?
I am doing a whole summary of the trigonometry chapter at the moment , so that i'll be more prepared for tha exam. These 3 equations were the only ones i couldn't solve , i asked help from my parents , my brother but they couldn't solve it either. I wasn't left with much choices so i went online and posted them here and i think this is the most adequate place to do it. No, i'm not asking someone else to do them for me . I just want a hint, a way of doing them so that i can learn something and put it to use later.

4. ## Re: trig equations

Originally Posted by enea54
1) sin2x + sinx * cosx = 1
$\displaystyle sin^2(x) + sin(x)~cos(x) = 1$

$\displaystyle (sin^2(x) -1) + sin(x)~cos(x) = 0$

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

I'd multiply both sides by -1, but you can do it from this too. Factor out the common cos(x) and the solution follows easily from there.

-Dan

5. ## Re: trig equations

thanks a lot , that was what i needed, thanks

6. ## Re: trig equations

For the second one you can replace 'x' with a new variable 'w' where x = 2w. This gives

$\displaystyle 1 + \cos(2w) + \cos(w) = 0$

Now use the identity $\displaystyle cos(2w)= 2cos^2w-1$ and solve for cos(w), and from that determine values for w, then x.

7. ## Re: trig equations

Ya beat me ebaines! Almost word for word.

-Dan

8. ## Re: trig equations

Notice that $\displaystyle sin^2(x) + sin(x)~cos(x) = 1$

can be written as $\displaystyle \cos(x)[\sin(x)-\cos(x)]=0$

9. ## Re: trig equations

Hello, enea54!

Here's the last one . . .

$\displaystyle (3)\;1 + \cos(x + \pi) + \cos\left(\tfrac{x}{2}+\tfrac{\pi}{2}\right) \:=\:0$

We have: .$\displaystyle 1 + \cos(x + \pi) + \cos\left(\tfrac{x+\pi}{2}\right) \:=\:0$

Let $\displaystyle \theta \:=\:x + \pi$

We have: .$\displaystyle 1 + \cos\theta + \cos\tfrac{\theta}{2} \:=\:0$

n . $\displaystyle 1 + \left(2\cos^2\!\tfrac{\theta}{2} - 1\right) + \cos\tfrac{\theta}{2} \:=\:0$

. . . . . . . . . $\displaystyle 2\cos^2\!\tfrac{\theta}{2} + \cos\tfrac{\theta}{2} \:=\:0$

n . . . . . $\displaystyle \cos\tfrac{\theta}{2}\left(2\cos\tfrac{\theta}{2} + 1\right) \:=\:0$

$\displaystyle \cos\tfrac{\theta}{2} \:=\:0 \quad\Rightarrow\quad \tfrac{\theta}{2} \:=\:\begin{Bmatrix}\frac{\pi}{2} \\ \\[-4mm] \frac{3\pi}{2} \end{Bmatrix} \quad\Rightarrow\quad \theta \:=\:\begin{Bmatrix} \pi \\ 3\pi \end{Bmatrix}$

. . . . $\displaystyle x + \pi \:=\:\begin{Bmatrix}\pi \\ 3\pi\end{Bmatrix} \quad\Rightarrow\quad \boxed{x \:=\:\begin{Bmatrix}0 \\ 2\pi\end{Bmatrix}}$

$\displaystyle 2\cos\tfrac{\theta}{2} + 1 \:=\:0 \quad\Rightarrow\quad \cos\tfrac{\theta}{2} \:=\:-\tfrac{1}{2} \quad\Rightarrow\quad \tfrac{\theta}{2} \:=\:\begin{Bmatrix}\frac{2\pi}{3} \\ \\[-4mm] \frac{4\pi}{3} \end{Bmatrix} \quad\Rightarrow\quad \theta \:=\:\begin{Bmatrix}\frac{4\pi}{3} \\ \\[-4mm] \frac{8\pi}{3} \end{Bmatrix}$

. . . . $\displaystyle x + \pi \:=\:\begin{Bmatrix}\frac{4\pi}{3} \\ \\[-4mm] \frac{8\pi}{3} \end{Bmatrix} \quad\Rightarrow\quad \boxed{x \:=\:\begin{Bmatrix} \frac{\pi}{3} \\ \\[-4mm] \frac{5\pi}{3} \end{Bmatrix}}$

10. ## Re: trig equations

thanks a lot guys, you are the best of the best , thanks again