Parametric equations find the points for which the gradient is 3

Question

The parametric equations of a curve are $$ x = t + \cos(t)$$$$ y= \ln(1+\sin(t))$$

where $-\frac {1}{2}\pi < t <\frac {1}{2}\pi$

$i) $ Show that $\frac{dy}{dx} = \sec (t)$

$ii)$ Hence find the x-coordinates of the points on the curve at which the gradient is equal to 3. Give your answers correct to 3 significant figures.

I did part $i$ how do I do part $ii$, I get $t = cos^{-1}\frac {1}{3}$, I plug that into the equation $ x = t + cos(t) $ and I get $x = 1.56$ which is correct, but how do I get the other answer which is $-0.898$?

Answer 1

I'll continue from your second last step

$$ sec(t) = 3 $$ which means

$$ sec(-t) = 3 $$ (sec(x) = sec(-x))

from here, you get

$$ t = sec^{-1}3 $$ or $$ t = -sec^{-1}3 $$

Hence you'll get $\ x = 0.56 or x= -0.896 $