Attention required!

     

$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 bởi next?


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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 $0le sin(x) le 1$ and $0le cos(x)le 1$.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 to $1$, & the other must be equal khổng lồ zero.From that information, I"m sure you can finish the solution.

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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


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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 $sin x$ and $cos x$ which holds if & only if

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

or

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

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