How do you solve 2cos x+3= sin2x on the interval [0,2π]?

Feb 21, 2015

It's impossible.

The functions sinus and cosine are functions that assume values in $\left[- 1 , 1\right]$, so the values of the first member are in $\left[1 , 4\right]$, but the second member still assumes values in $\left[- 1 , 1\right]$ (the $2 x$, argument of the sinus, doesn't change the interval).

The only way to have a solution is that it would be $1$, the only one common value, but:

$\sin 2 x = 1 \Rightarrow 2 x = \frac{\pi}{2} + 2 k \pi \Rightarrow x = \frac{\pi}{4} + k \pi$

so the possibilities are: $\frac{\pi}{4} , \frac{5}{4} \pi$.

But the first member in $\frac{\pi}{4}$ assumes value:

$2 \cos \left(\frac{\pi}{4}\right) + 3 = 2 \frac{\sqrt{2}}{2} + 3 = \sqrt{2} + 3$, that isn't $1$,

and in $\frac{5}{4} \pi$:

$2 \cos \left(\frac{5}{4} \pi\right) + 3 = 2 \left(- \frac{\sqrt{2}}{2}\right) + 3 = - \sqrt{2} + 3$, that isn't $1$.

So there are no solutions!

$y = 2 \cos x + 3 - \sin 2 x$ and seeing that it doesn't touch the x-axis.