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

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

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Answer 1 · 2015-02-21T20:53:07

It's impossible.

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

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

$sin 2 x = 1 \Rightarrow 2 x = \frac{π}{2} + 2 k π \Rightarrow x = \frac{π}{4} + k π$ $sin2x=1rArr2x=pi/2+2kpirArrx=pi/4+kpi$

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

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

$2 cos (\frac{π}{4}) + 3 = 2 \frac{\sqrt{2}}{2} + 3 = \sqrt{2} + 3$ $2cos(pi/4)+3=2sqrt2/2+3=sqrt2+3$ , that isn't $1$ $1$ ,

and in $\frac{5}{4} π$ $5/4pi$ :

$2 cos (\frac{5}{4} π) + 3 = 2 (- \frac{\sqrt{2}}{2}) + 3 = - \sqrt{2} + 3$ $2cos(5/4pi)+3=2(-sqrt2/2)+3=-sqrt2+3$ , that isn't $1$ $1$ .

So there are no solutions!

You can see also looking the graph of the function:

$y = 2 cos x + 3 - sin 2 x$ $y=2cosx+3-sin2x$ and seeing that it doesn't touch the x-axis.

graph{2cosx+3-sin(2x) [-10, 10, -5, 5]}

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

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