Trigonometric Equation

Geometry Level 2

How many integers x such that cot x + tan x = 1

infinitely many solution 2 1 0 3

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Paul Ryan Longhas
Jan 24, 2015

cot x + tan x = 1 => (cos x / sin x) + (sin x/cos x) = {(sin x)^2+(cosx)^2}/sinxcosx = 1/sinx cosx => 2 / sin 2x = 1 => sin 2x = 2. But the range of sine function is [-1,1]. Therefore, there are no solution in this problem.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. Another approach would be to observe that by the A.M.-G.M. inequality, since cot ⁡ ( x ) = 1 tan ⁡ ( x ) \cot(x) = \frac{1}{\tan(x)} , we can conclude that

( cot ⁡ ( x ) + tan ⁡ ( x ) ) ≥ 2 (\cot(x) + \tan(x)) \ge 2 for tan ⁡ ( x ) > 0 \tan(x) \gt 0 and

( cot ⁡ ( x ) + tan ⁡ ( x ) ) ≤ − 2 (\cot(x) + \tan(x)) \le -2 for tan ⁡ ( x ) < 0 \tan(x) \lt 0 .

(For either cot ⁡ ( x ) = 0 \cot(x) = 0 or tan ⁡ ( x ) = 0 \tan(x) = 0 the sum would be indeterminate.)

Thus cot ⁡ ( x ) + tan ⁡ ( x ) \cot(x) + \tan(x) can never equal any value on the interval ( − 2 , 2 ) . (-2,2).

Brian Charlesworth - 6 years, 4 months ago

tan(x)+cot(x) has minimum value of 2! so there is no chance of the trigonometric function becoming equal to 1!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...