Question

I had this trigonometric equation at math competition and Idk how to solve it

Answer 1

\[\sin^{1995}x+\cos^{1995}x=1\]

Answer 2

-1<=sinx<=1
\[\LARGE -1 \leq \sin^{1995}x \leq 1\]
\[\LARGE -1 \leq \cos^{1995}x \leq 1\]
since 1995 is odd..
sinx = -1,1,0,1/2
cosx = -1,1,0,1/2
=1 when
Case 1 : sinx=1 , cosx=0
Case 2 : sinx=0, cosx=1
Case 3 : sinx=1/2 , cosx=1/2
don't know if im going right,we have insufficient information,do u have the answer?

Answer 3

no I don't :(

Answer 4

Discard case 3,
x=0,1 (answer)

Answer 5

Case 3 is not possible since both sinx and cosx cant be 1/2 simultaneously,either of them has to yield its max value,hence
x=0,1=answer

Answer 6

@DLS @Fificol

Using the two cases that DLS mentioned, namely when sin(x) = 0 and cos(x) = 1 and cos(x) = 1 and sin(x) = 0, then we get the following solutions:\[\bf x = 2\pi k \ \ or \ \ x=\frac{\pi}{2}+2\pi k \ | \ k \in \mathbb{Z}\]

Answer 7

yup

Answer 8

I think these are only two solutions because of the very norrow shape of sin and cos
Thanks for help:)