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Solutions to an equation (B.Stat 2005 Subjective Problem 4 Solution)

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

Discussion:

Teacher: Notice that \(|\sin x| \le 1 , |\cos x | \le 1 \) . So if you raise sin x and cos x to higher powers you necessarily lower the value. Take for example the number 1/2. If you raise that to the power of 2, you get 1/4. Raise it to the power of 3, you get 1/8 (which is smaller than 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 \(sin ^5 x \le |\sin x|^5 \le \sin^2 x , \cos^3 x \le |\cos x |^3 \le \cos^2 x \). Adding these two inequalities we get \(\sin ^5 x + \cos^2 x \le \sin^2 x + cos^2 x = 1 \)

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

\(\sin ^{5} x = \sin ^{3} x implies \sin^{3} x (\sin^{2} x – 1 ) = 0 implies \sin^{3} x = 0 \text{or} \sin^{2} x = 1 \)

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

Similarly \(\cos ^{3} x = \cos ^{2} x implies \cos^{2} x (\cos x – 1 ) = 0 implies \cos^{2} x = 0 \) 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.

May 7, 2015

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