Solve-for-real-numbers-sin-sinx-sinx-cos-cosx-cosx-1-

Question Number 138725 by mathsuji last updated on 17/Apr/21

S o l v e f o r r e a l n u m b e r s

\frac{s i n (s i n x)}{s i n x} + \frac{c o s (c o s x)}{c o s x} = 1

Answered by mr W last updated on 18/Apr/21

t = \sin x

\cos x = \pm \sqrt{1 - t^{2}}

- 1 ⩽ t ⩽ 1

f (t) = \frac{\sin t}{t} + \frac{\cos \sqrt{1 - t^{2}}}{\sqrt{1 - t^{2}}} ⩾ f (0) = 1 + \cos 1 > 1

\Rightarrow n o s o l u t i o n f o r f (t) = 1

g (t) = \frac{\sin t}{t} - \frac{\cos \sqrt{1 - t^{2}}}{\sqrt{1 - t^{2}}} ⩽ g (0) = 1 - \cos 1 < 1 =

\Rightarrow n o s o l u t i o n f o r g (t) = 1

\Rightarrow n o s o l u t i o n f o r

\frac{s i n (s i n x)}{s i n x} + \frac{c o s (c o s x)}{c o s x} = 1!

\frac{s i n (s i n x)}{s i n x} + \frac{c o s (c o s x)}{c o s x} = a h a s r o o t s

o n l y i f a ⩾ 1 + \cos 1 o r a ⩽ 1 - \cos 1

Commented by mathsuji last updated on 20/Apr/21

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