Find-the-real-roots-of-the-equation-cos-4-x-sin-7-x-1-in-the-interval-pi-pi-

Question Number 14977 by Tinkutara last updated on 06/Jun/17

Find the real roots of the equation

\cos^{4} x + \sin^{7} x = 1 in the interval [- π, π] .

Commented by mrW1 last updated on 06/Jun/17

\cos x = \pm 1 a n d \sin x = 0

o r

\cos x = 0 a n d \sin x = 1

\Rightarrow x = - π, 0, \frac{π}{2}, π

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

Answered by myintkhaing last updated on 07/Jun/17

{s i n}^{7} x = 1 - {c o s}^{4} x

{s i n}^{7} x = {s i n}^{2} x (2 - {s i n}^{2} x)

{s i n}^{2} x ({s i n}^{5} x + {s i n}^{2} x - 2) = 0

{s i n}^{2} x = 0 o r {s i n}^{5} x + {s i n}^{2} x - 2 = 0

s i n x = 0 \Rightarrow x = 0, \pm π

{s i n}^{5} x + {s i n}^{2} x - 2 = 0, h e r e - 1 ⩽ s i n x ⩽ 1

i f s i n x = - 1, L . H . S \neq R . H . S

i f s i n x = 1, L . H . S = R . H . S

x = \frac{π}{2}

x = 0, \pm π, \frac{π}{2}

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir! But why you reject the 2^{nd}

factor ?

Commented by myintkhaing last updated on 07/Jun/17

M y p o s t i s e d i t e d .

Commented by Tinkutara last updated on 07/Jun/17

But how to identify this that it has no

real root ?

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

Answered by mrW1 last updated on 07/Jun/17

This question can be generalised to

\cos^{2 m} x + \sin^{2 n + 1} x = 1 with m, n \in N^{+}

the answer is the same, independent

from m and n .

\sin^{2 n + 1} x = 1 - \cos^{2 m} x ⩾ 0 and ⩽ 1

\Rightarrow 0 ⩽ \sin x ⩽ 1

for \sin x = 0 we get \cos^{2 m} x = 1 or

\cos x = \pm 1, this is possible .

for \sin x = 1 we get \cos^{2 m} x = 0 or

\cos x = 0, this is possible .

for 0 < \sin x < 1 we get

\cos^{2 m} x + \sin^{2 n + 1} x = 1

{(\cos^{2} x)}^{m} + \sin^{2 n + 1} x = 1

{(1 - \sin^{2} x)}^{m} + \sin^{2 n + 1} x = 1

let t = \sin x, 0 < t < 1

\Rightarrow {(1 - t^{2})}^{m} + t^{2 n + 1} = 1

{(1 - t^{2})}^{m} = 1 - t^{2 n + 1} = (1 - t) (1 + t + t^{2} + \dots + t^{2 n})

\Rightarrow \frac{{(1 - t^{2})}^{m}}{1 - t} = 1 + t + t^{2} + \dots + t^{2 n} > 1 + t

\Rightarrow \frac{{(1 - t^{2})}^{m}}{(1 - t) (1 + t)} > 1

\Rightarrow {(1 - t^{2})}^{m - 1} > 1

but 0 < 1 - t^{2} < 1 \Rightarrow {(1 - t^{2})}^{m - 1} < 1

this is a contradiction, which results

from the assumption that 0 < \sin x < 1 .

i . e . the assumption is wrong .

therefore the solution of the equation

\cos^{2 m} x + \sin^{2 n + 1} x = 1 with m, n \in N^{+}

is only

\sin x = 0 and \cos x = \pm 1

\Rightarrow x = k π, k \in Z

or

\sin x = 1 and \cos x = 0

\Rightarrow x = k π + {(- 1)}^{k} \frac{π}{2}, k \in Z

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!