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May 7, 2015

Solutions to an equation | B.Stat 2005 Subjective Problem 4

Problem:

Find all real solutions of the equation $sin^{5}x+cos^{3}x=1$ .

Discussion:

Teacher: Notice that $|\sin x| \leq 1 , |\cos x | \leq 1 $ . So if you raise $\sin x$ and $\cos x$ to higher powers you necessarily lower the value. Take for example the number $\frac{1}{2}$. If you raise that to the power of $2$, you get $\frac{1}{4}$. Raise it to the power of $3$, you get $\frac{1}{8}$ (which is smaller than $\frac{1}{4}$).
When absolute value of a number is greater than $1$, values increase, with increasing power. If the absolute value is less than $1$, the values decrease with increasing power.
Use this insight to solve this problem.

Student: I see. Then $sin ^5 x \leq |\sin x|^5 \leq \sin^2 x , \cos^3 x \leq |\cos x |^3 \leq \cos^2 x $. Adding these two inequalities we get $\sin ^5 x + \cos^2 x \leq \sin^2 x + cos^2 x = 1 $

Equality holds when simultaneously $\sin^5 x = \sin^2 x $,
$\cos^3 x = \cos^2 x $

$\sin ^{5} x = \sin ^{3} x \Rightarrow \sin^{3} x (\sin^{2} x - 1 ) = 0 \Rightarrow \sin^{3} x = 0$ or $\sin^{2} x = 1 $

This implies $x = n \pi $ or $ \displaystyle {x = (2n+1) \frac{\pi}{2} } $

Similarly $\cos ^{3} x = \cos ^{2} x \Rightarrow \cos^{2} x (\cos x - 1 ) = 0 \Rightarrow \cos^{2} x = 0 $ or $\cos x = 1 $

This implies $x = 2n \pi $ or $ \displaystyle {x = (2n+1) \frac{\pi}{2} } $

Therefore simultaneously $\sin^5 x = \sin^2 x , \cos^3 x =\cos^2 x $ holds when $x = 2n \pi $ or $\displaystyle {x = (2n+1) \frac{\pi}{2} } $ where $n$ is any natural number.

Some Useful Links:

Solving a few Diophantine Equations – Video

ISI 2015 BStat – BMath Objective Problems

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