With the domain of (-pi/4<=pi/4), how is ln|secx+tanx|=ln|3+2sqrt(2)| ?

Answer 1

Let y = ln l secx + tanx l. y is continuous on the interval [ -π/4, π/4 ].

So, y achieves maximum and minimum values on the interval.

dy/dx = (secxtanx + sec²x) / (secx + tanx) = secx(tanx + secx) / (secx + tanx) = secx = 1 / cosx

dy/dx > 0 on the interval [-π/4, π/4]. So y is increasing on the interval,.

Therefore, the maximum value of y on the interval occurs at the right endpoint and the minimum value of y occurs at the left endpoint.

Maximum value of y = ln l secx + tanx l on the interval is ln l sec(π/4) + tan (π/4) l = ln (√2 + 1) ≈ 0.88

However, ln (3 + 2√2) ≈ 1.76

So, the equation ln l secx + tanx l = ln(3 + 2√2) has NO SOLUTION on the interval [-π/4, π/4 ].

Answer 2

Let's assume that this equation does have a solution. Then we can write it as

sec(x) + tan(x) = +/- (3 + 2sqrt(2))

Writing everything in terms of sin(x) and cos(x) gives us

1 / cos(x) + sin(x) / cos(x) = +/- (3 + 2sqrt(2))

Multiplying both sides by cos(x), we get

1 + sin(x) = +/- (3 + 2sqrt(2)) cos(x)

Squaring both sides...

1 + 2 sin(x) + sin(x)^2 = (17 + 12sqrt(2)) cos(x)^2

Replacing cos(x)^2 with 1 - sin(x)^2...

1 + 2 sin(x) + sin(x)^2 = (17 + 12sqrt(2)) (1 - sin(x)^2)

Subtracting the right hand side from both sides of the equation and collecting terms...

(18 + 12sqrt(2) sin(x)^2 + 2 sin(x) + (-16 - 12sqrt(2)) = 0

This is a quadratic in sin(x). Plugging these coefficients into the quadratic formula yields...

sin(x) = (-2 +/- sqrt(4 - 4(18 + 12sqrt2)(-16 - 12sqrt(2))) / (2(18 + 12sqrt(2))

Which simplifies to...

sin(x) = (-1 +/- sqrt(577 + 408sqrt(2))) / (18 + 12sqrt(2))

When we compute this result by choosing the minus sign, we get sin(x) = -1 which is true when x = -pi/2. This gives a division by 0 when plugged back into the original equation, and is also outside the domain of the problem statement, so we can discard x = -pi/2.

When we compute the result by choosing the positive sign, we get...

sin(x) = (-1 + sqrt(577 + 408sqrt(2))) / (18 + 12sqrt(2))

And so by using the inverse sine function on both sides, we get

x = arcsin(-1 + sqrt(577 + 408sqrt(2))) / (18 + 12sqrt(2)) = 1.230959417...

We also know that pi minus the above number is also a solution, so the two solutions are...

x = 1.230959417... + 2k*pi, x = 1.910633236... + 2k*pi, for any integer k.

These numbers for x indeed solve the original equation, however they fall outside the desired domain, thus we conclude that there are no solutions to this problem.

Answer 3

-45 to 45 degrees

for secx x cannot = 90 or -90, which is irrelevant

for tanx x cannot = 90 or -90 which is irrelevant

given the restriction -45<x</=45 or -pi/4 <x</= pi/4

but logs or natural logs have to have positive x

secx has to be > tanx or would have, except for the absolute value signs

also

secx+tanx has to be= 3+2sqr2= 3+2^1.5 = about 3+2.82 =5.82

x cannot = -5, 7 or 5/4 as then secx+tanx is undefined

x is in the union of intervals (-pi/4, 1.25)U(1.25, pi/4)

one method is use a graphing calculator and see the relevant values of x

use desmos or TI83