# Just A Moment

\$sin^3x+cos^3x=1\(sin x+cos x)(sin^2x-sin xcdotcos x+cos^2x)=1\(sin x+cos x)(1-sin xcdotcos x)=1\$

What should I do next?

Bạn đang xem: Just a moment   Both of \$sin(x),cos(x)\$ must be nonnegative since, if one of them was negative, the equation \$sin^3(x)+cos^3(x)=1\$ would imply that the other one is more than \$1\$, contradiction.Thus, we have \$0le sin(x) le 1\$ & \$0le cos(x)le 1\$.If both of \$sin(x),cos(x)\$ are less than \$1\$, then, since they are both nonnegative, we would have \$\$egincases0le sin^3(x) but that would imply\$\$sin^3(x)+cos^3(x) contradiction.It follows that one of \$sin(x),cos(x)\$ must be equal khổng lồ \$1\$, và the other must be equal to lớn zero.From that information, I"m sure you can finish the solution. Xem thêm: Trình Bày Đặc Điểm Cấu Tạo Ngoài Của Thằn Lằn, Thích Nghi Với Đời Sống Hoàn Toàn Ở Cạn

Hint:

Let \$sin x+cos x=timplies t^2=?\$

\$\$1=dfract2-(t^2-1)2iff t^3-3t+2=0\$\$

Clearly, \$t=1\$ a solution Let use by \$t= an (x/2)\$

\$sin x=frac2t1+t^2\$

\$cos x=frac1-t^21+t^2\$

to obtain

\$\$2t^6-8t^3+6t^2=0\$\$

\$\$iff t^2(t^4-4t+3)=t^2(t-1)^2(t^2+2t+3)=0\$\$

and since \$t^2+2t+3>0\$ the solutions are

\$t=0 implies frac x 2=kpiimplies x=2kpi\$

\$t=1 implies frac x 2=fracpi4+kpi implies x=fracpi2+2kpi \$

As an alternative by

\$\$sin^3 x+ cos ^3 x=1 iff sin xcdot sin^2+cos xcdot cos^2 x=1\$\$

since \$sin^2 x+ cos ^2 x=1\$, the given equality is a weighted mean of \$sin x\$ & \$cos x\$ which holds if và only if

\$sin x=1,,cos x=0\$

or

\$cos x=1,,sin x=0\$

Xem thêm: Mục Đích Của Việc Sử Dụng Biểu Đồ Là Gì? Các Loại Biểu Đồ

Site thiết kế / logo © 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. Rev2022.12.19.43125