## IB Maths SL Trigonometric equations

Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths SL Revision Notes

### IB Maths SL Trigonometric equations

IB Mathematics SL– Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

$2cos^2x=cosx+1$ for $- \pi \leq x \leq \pi$

Thanks
Jane

Posts: 0
Joined: Tue Mar 19, 2013 8:47 pm

### Re: IB Maths SL Trigonometric equations

IB Maths SL – Trigonometry, Trigonometric equations

By treating the equation as a quadratic in $cosx$ and then factoring.

$2cos^2x=cosx+1\Leftrightarrow 2cos^2x-cosx-1=0 \Leftrightarrow$

$\Leftrightarrow (2cosx+1)(cosx-1)=0 \Leftrightarrow cosx-1=0 \ or \ 2cosx+1 =0 \Leftrightarrow$

Now, you have to solve two trigonometric equations

$cosx-1=0$ and $2cosx+1 =0$ in the interval $[-\pi , \pi]$

For the first equation we have:

$cosx-1=0\Leftrightarrow cosx=1$

The above equation has only one solution in the interval $[-\pi , \pi]$

$x=0$

For the second equation we have:

$2cosx+1 =0 \Leftrightarrow cosx=-\frac{1}{2}$

The above equation has two solutions in the interval $[-\pi , \pi]$

$x=-\pi +\frac{\pi}{3}=-\frac{2\pi}{3} \ or\ x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: IB Maths SL Trigonometric equations

Thanks a lot!
Jane

Posts: 0
Joined: Tue Mar 19, 2013 8:47 pm