1. ## triq eqautions

cos2t+2cos^2(t/2)-2=0

2. Originally Posted by Thelema
cos2t+2cos^2(t/2)-2=0
$\cos 2t+2\cos^2(\frac{t}2)-2=0$

$\cos 2t+(2\cos^2(\frac{t}2)-1) -1=0$

$\cos 2t+\cos t -1=0$

$2\cos^2 t - 1 +\cos t -1=0$

$2\cos^2 t +\cos t - 2=0$

I am sure you know what to do now

3. Originally Posted by Thelema
cos2t+2cos^2(t/2)-2=0
Double angle formula: $\cos (2A) = 2 \cos^2 (A) - 1$.

Substitute A = t/2: $\cos (t) = 2 \cos^2 (t/2) - 1 \Rightarrow \cos^2 (t/2) = \frac{1}{2} \, (1 + \cos t)$.

So your equation can be re-written as

$(2 \cos^2 t - 1) + (1 + \cos t) - 2 = 0 \Rightarrow 2 \cos^2 t + \cos t - 2 = 0$ ......

Check for mistakes - but you get the idea, right?

4. $\cos 2t + 2\cos^2 \frac{t}{2} - 2 = 0$

$\cos 2t + 2\cos^2 \frac{t}{2} = 2$

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

..... $\cos^2 \frac{t}{2} = \frac{1+\cos t}{2}$.....

$\cos^2 t - \sin^2 t + \not{2}.\frac{1+\cos t}{\not 2} = 2$

$\cos^2 t - \sin^2 t + 1 + \cos t = 2$

..... $1 = \cos^2 t + \sin^2 t$.....

$2\cos^2 t + \cos t = 2$

Let $a= \cos t$..

$2a^2+ a -2 = 0$

$a = \cos t = \frac{\sqrt{17}-1}{4},\frac{-\sqrt{17}-1}{4}$

$\cos t = \frac{-\sqrt{17}-1}{4}$ is not possible because cos is between -1 and +1.

So we only use
$\cos t = \frac{\sqrt{17}-1}{4}$

Solve this for t.