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by **Nyla** » Sun Aug 18, 2013 10:20 am

MYP Mathematics Pre-diploma – Trigonometry, Trigonometric equations

How can we solve the following trigonometric equation?

$2cos^2x=cosx+1$ for $0 \leq x \leq 2\pi$

Thanks

by **Jack** » Sun Aug 18, 2013 10:26 am

MYP Mathematics Pre-diploma – 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 $[0 , 2\pi]$

For the first equation we have:

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

The above equation has two solutions in the interval $[0, 2\pi]$

$x=0 \ or \ x=2\pi$

For the second equation we have:

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

The above equation has two solutions in the interval $[0 , 2\pi]$

$x=\pi +\frac{\pi}{3}=\frac{4\pi}{3} \ or\ x=2\pi-\frac{\pi}{3}=\frac{5\pi}{3}$

Hope these help!!

by **Nyla** » Sun Aug 18, 2013 10:30 am

Thanks Jack.

ib maths4u | ib maths | ib math | ib mathematics

MYP math Trigonometric equations

MYP math Trigonometric equations

Re: MYP math Trigonometric equations

Re: MYP math Trigonometric equations