Matematické Fórum

stuart clark · 10. 06. 2014 08:28

Calculation of all real values of $x$ in $\sin^5 x+\cos^3 x = 1$

sugyman · 13. 06. 2014 17:33

Using Pythagorean identity:
$\sin^5 x+\cos^3 x = \sin^2x+\cos^2x$
$\sin^2 x(\sin^3x-1)+\cos^2 x(\cos x -1)=0$
since $\sin^2 x(\sin^3x-1)\leq 0$ and $\cos^2 x(\cos x -1)\leq0$ , then:
$\sin^2 x(sin^3x-1)=0$ and $\cos^2 x(\cos x -1)$ .
If $\sin^2x=0 \Rightarrow x=k\pi$ also $\cos^2x=1$ , $\cos x-1$ has to be therefore zero $\Rightarrow \cos x=1 \Rightarrow x=2k\pi$ .Overall: $x=2k\pi$
And the second case:
$\sin^3x-1=0 \Rightarrow \sin x=1 \Rightarrow x=\frac{\pi}{2}+2k\pi$ and $\cos^2x=0 \Rightarrow \cos x=\frac{\pi}{2}+k\pi$ . Overall: $x=\frac{\pi}{2}+2k\pi$ .
To sum up, solutions: $x=\frac{\pi}{2}+2k\pi$ and $x=2k\pi$ for whole k.

stuart clark · 14. 06. 2014 06:24

Thanks ↑ sugyman:

Matematické Fórum

#1 10. 06. 2014 08:28

Trig. equation

#2 13. 06. 2014 17:33

Re: Trig. equation

#3 14. 06. 2014 06:24

Re: Trig. equation

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