Exponential Trigonometry!

Geometry Level 4

$\Large{2^{\sin(x) - \cos(x)} = \tan(x)}$

Find the sum of all values of $x$ such that $x \in (0, 2\pi)$ satisfying the above equation upto three correct places of decimals.

Note: $x$ is measured in radians.

The answer is 4.712.

2 solutions

Chew-Seong Cheong
Sep 7, 2015

$\begin{aligned} 2^{\sin{x}-\cos{x}} & = \tan{x} \\ \Rightarrow \frac{2^{\sin{x}}}{2^{\cos{x}}} & = \frac{\sin{x}}{\cos{x}} \\ \frac{\sin{x}}{2^{\sin{x}}} & = \frac{\cos{x}}{2^{\cos{x}}} \\ \Rightarrow \sin{x} & = \cos{x} \\ \Rightarrow x & = \begin{cases} \frac{\pi}{4} \\ \frac{5\pi}{4} \end{cases} \text{for } x \in (0, 2\pi) \end{aligned}$

The required answer is $\frac{\pi}{4} + \frac{5\pi}{4} = \frac{3\pi}{2} \approx \boxed{4.712}$

its 4.712 not 4.172

Soundarya Gupta - 5 years, 9 months ago

Thanks. I have changed it.

Chew-Seong Cheong - 5 years, 9 months ago

Note that you have to show that $f: x \mapsto \frac{x}{2^x}$ is bijective.

Jake Lai - 5 years, 7 months ago

It only needs to be injective, not necessarily bijective.

Wanchun Shen - 5 years, 7 months ago

Kenny Lau
Sep 9, 2015

Transform the equation into $\dfrac{\sin x}{2^{\sin x}}=\dfrac{\cos x}{2^{\cos x}}$ .

An analysis of the function $f(x)=\dfrac{x}{2^x}$ would tell us that the function is increasing when $-1\le x\le1$ .

Therefore, $f(x)=f(y)\implies x=y$ when $-1\le x,y\le1$ .

Therefore, $\sin x=\cos x$ , which gives us $x=\dfrac\pi4$ or $x=\dfrac{5\pi}4$ .

Exponential Trigonometry!

The answer is 4.712.

2 solutions

0 pending reports