+0

coshX*cosx=0 x=?

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coshX*cosx=0  x=?

Guest Apr 15, 2017
#1
+1

Solve for x over the real numbers:
cos(x) cosh(x) = 0

Split into two equations:
cos(x) = 0 or cosh(x) = 0

Take the inverse cosine of both sides:
x = π/2 + π n for n element Z
or cosh(x) = 0

cosh(x) = 0 has no solution since for all x element R, cosh(x)>=1 and True:
Answer: | x = π/2 + π n for n element Z

Guest Apr 15, 2017
#3
+7489
0

Hi Guest, your solution for coshx * cosx = 0 is not correct.

There are only four results, not infinitely many.

Greeting! asinus

Oh sorry, I've seen this wrong.

Your solution for coshx * cosx = 0 is correct.

I am sorry! asinus

asinus  Apr 16, 2017
edited by asinus  Apr 17, 2017
#2
+7489
+1

coshx*cosx=0  x=?

$$cos\ x =0\\x=\frac{\pi}{2}+\pi\cdot n\\n \in ℤ$$

$$cosh\ x=\frac{1}{2}(e^x+e^{-x})>0\\ for \ x\in \Re$$    no solution

$$solution:$$

$$coshx \cdot cosx=0\\x_1=\frac{\pi}{2} \\x_2=\frac{\pi}{2}+\pi\\x_3=-\frac{\pi}{2}\\x_4=-(\frac{\pi}{2}+\pi)$$

$$x=\frac{\pi}{2}+\pi\cdot n$$     Thank you! C.

$$for\ n \in ℤ$$

!

asinus  Apr 16, 2017
edited by asinus  Apr 16, 2017
edited by asinus  Apr 17, 2017
#4
+90001
+2

Actually, the guest is correct....this graph is symmetric around the y axis  and will = 0 at

pi/2 + n*pi      radians......where n is an integer.......

See the graph below :

CPhill  Apr 16, 2017
#5
+7489
+1

Oh sorry, I've seen this wrong. I am sorry! asinus

asinus  Apr 16, 2017