1. ## solve equation

1.solve for x in $sinx tanx+ tanx-2sinx+cosx=0$ for 0<orequalx<or equal $2\pi$ rads

2. find the exact value of $(\frac{7\pi}{6})$

2. Here is the graph of the function in the left side of the equation.

The equation has one solution between 3 and 4.

(1)
sin^2 x +cos^2 x + sin x -2 sin x cos x =0

sin x (1- 2 cos x )=0

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

Hello william

What do you mean? This doesn't make sense.
Originally Posted by william
...
2. find the exact value of $(\frac{7\pi}{6})$
If you mean 'Find the exact value of $\sin(\tfrac{7\pi}{6})$', then the answer is that $\sin(\pi+A) = -\sin A$, and $\sin(\tfrac{\pi}{6}) = \tfrac12$. So...?

6. Originally Posted by red_dog
Here is the graph of the function in the left side of the equation.

The equation has one solution between 3 and 4.
By a numerical method, I got $x=3.4968$, correct to 4 d.p. Are you sure we have the correct question here?

7. I tried another way, but that doesn't lead to an exact solution.

$\sin x-2\sin x\cos x+1=0$

$2\sin\frac{x}{2}\cos\frac{x}{2}\left(\sin^2\frac{x }{2}+\cos^2\frac{x}{2}\right)-$

$-4\sin\frac{x}{2}\cos\frac{x}{2}\left(\cos^2\frac{x }{2}-\sin^2\frac{x}{2}\right)+\left(\sin^2\frac{x}{2}+\ cos^2\frac{x}{2}\right)^2=0$

$\sin^4\frac{x}{2}+6\sin^3\frac{x}{2}\cos\frac{x}{2 }+2\sin^2\frac{x}{2}\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos^3\frac{x}{2}+\cos^4\frac{x}{2 }=0$

Divide by $\cos^4\frac{x}{2}$ and let $\tan\frac{x}{2}=t$

$t^4+6t^3+2t^2-2t+1=0$

$(t+1)(t^3+5t^2-3t+1)=0$

$t=-1$ which is not good because $x\neq\frac{(2k+1)\pi}{2}$

$t^3+5t^2-3t+1=0$ which has an irational solution.