1. ## trigonometry proof help.

2cosx - secx = - tanx

i need to solve the equation and give an answer in radians.

2. The equation you wrote was:

$\displaystyle 2cos(x)-sec(x)=-tan(x)$

I'm not sure if this is true. Consider the angle $\displaystyle \frac{\pi}{3}$

$\displaystyle 2cos\left(\frac{\pi}{3}\right)-sec\left(\frac{\pi}{3}\right)=-tan\left(\frac{\pi}{3}\right)$

The left side becomes:
$\displaystyle 2(.5)-2=-1$

The right side becomes:

$\displaystyle tan(\frac{\pi}{3})=\sqrt{3}\approx 1.73$

3. Hello starrytulips143
Originally Posted by starrytulips143
2cosx - secx = - tanx

i need to solve the equation and give an answer in radians.
Here are the steps to take; I'll leave the details to you:

• Using $\displaystyle \sec x = \frac{1}{\cos x}$ and $\displaystyle \tan x = \frac{\sin x}{\cos x}$, express everything in terms of sine and cosine.

• Multiply both sides by $\displaystyle \cos x$ to get rid of fractions.

• Then use $\displaystyle \cos^2x+\sin^2x=1$ to express everything in terms of $\displaystyle \sin x$.

• Re-arrange as a quadratic in $\displaystyle \sin x$.

• Factorise. You should get:

$\displaystyle (2\sin x+1)(\sin x -1)=0$
Can you complete it now?