Menu Close

Find-all-values-of-x-y-and-k-for-which-the-system-of-equations-sin-x-cos-2y-k-4-2k-2-2-cos-x-sin-2y-k-1-has-a-solution-




Question Number 18893 by Tinkutara last updated on 01/Aug/17
Find all values of x, y and k for which  the system of equations  sin x cos 2y = k^4  − 2k^2  + 2  cos x sin 2y = k + 1  has a solution.
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x},\:{y}\:\mathrm{and}\:{k}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\mathrm{sin}\:{x}\:\mathrm{cos}\:\mathrm{2}{y}\:=\:{k}^{\mathrm{4}} \:−\:\mathrm{2}{k}^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$$\mathrm{cos}\:{x}\:\mathrm{sin}\:\mathrm{2}{y}\:=\:{k}\:+\:\mathrm{1} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{solution}. \\ $$
Answered by 433 last updated on 01/Aug/17
     { ((sin xcos 2y=(k^2 −1)^2 +1)),((cos xsin 2y=k+1)) :}  −1≤k+1≤1 & −1≤(k^2 −1)^2 +1≤1  −2≤k≤0 & −2≤(k^2 −1)^2 ≤0  −2≤k≤0 & k=±1  k=−1   { ((sin xcos 2y=0)),((cos xsin 2y=0)) :}  sin x=0 ⇒cos x=±1  cos 2y=0 ⇒ sin 2y=±1  (x,y)={(2kπ,mπ),(2kπ,mπ+(π/2)),(2kπ+π,mπ),(2kπ+π,mπ+(π/2)),(2kπ±(π/2),mπ±(π/4))}
$$ \\ $$$$ \\ $$$$\begin{cases}{\mathrm{sin}\:{x}\mathrm{cos}\:\mathrm{2}{y}=\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\\{\mathrm{cos}\:{x}\mathrm{sin}\:\mathrm{2}{y}={k}+\mathrm{1}}\end{cases} \\ $$$$−\mathrm{1}\leqslant{k}+\mathrm{1}\leqslant\mathrm{1}\:\&\:−\mathrm{1}\leqslant\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\leqslant\mathrm{1} \\ $$$$−\mathrm{2}\leqslant{k}\leqslant\mathrm{0}\:\&\:−\mathrm{2}\leqslant\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$−\mathrm{2}\leqslant{k}\leqslant\mathrm{0}\:\&\:{k}=\pm\mathrm{1} \\ $$$${k}=−\mathrm{1} \\ $$$$\begin{cases}{\mathrm{sin}\:{x}\mathrm{cos}\:\mathrm{2}{y}=\mathrm{0}}\\{\mathrm{cos}\:{x}\mathrm{sin}\:\mathrm{2}{y}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{sin}\:{x}=\mathrm{0}\:\Rightarrow\mathrm{cos}\:{x}=\pm\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{2}{y}=\mathrm{0}\:\Rightarrow\:\mathrm{sin}\:\mathrm{2}{y}=\pm\mathrm{1} \\ $$$$\left({x},{y}\right)=\left\{\left(\mathrm{2}{k}\pi,{m}\pi\right),\left(\mathrm{2}{k}\pi,{m}\pi+\frac{\pi}{\mathrm{2}}\right),\left(\mathrm{2}{k}\pi+\pi,{m}\pi\right),\left(\mathrm{2}{k}\pi+\pi,{m}\pi+\frac{\pi}{\mathrm{2}}\right),\left(\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{2}},{m}\pi\pm\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$
Commented by Tinkutara last updated on 01/Aug/17
Thank you very much 433 Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{433}\:\mathrm{Sir}! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *