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Secant(x) + cosine(x) = 1

 Jan 7, 2017

Best Answer 

 #4
avatar+37091 
+5

Well, I should have specified no REAL solution.....

Yikes...you DO like those long calcs don't you ???

Thanx !!

 Jan 7, 2017
 #1
avatar+37091 
+5

I may be wrong, but when I graph this I see no place where this equation would have a solution....   Anyone else?

Hope I am wrong and can learn something (like you!)  .....

 Jan 7, 2017
 #2
avatar+37091 
+5

....another thought...the secant and cosine of x are always of the same sign  (sec = 1/cos )   so for this equation to be answered they both need to be POSITIVE.    Now Secant is ALWAYS >= 1     so this question has no solution!

ElectricPavlov  Jan 7, 2017
 #3
avatar
+5

Solve for x:
cos(x) + sec(x) = 1

Multiply both sides of cos(x) + sec(x) = 1 by cos(x):
1 + cos^2(x) = cos(x)

Subtract cos(x) from both sides:
1 - cos(x) + cos^2(x) = 0

Subtract 1 from both sides:
cos^2(x) - cos(x) = -1

Add 1/4 to both sides:
1/4 - cos(x) + cos^2(x) = -3/4

Write the left hand side as a square:
(cos(x) - 1/2)^2 = -3/4

Take the square root of both sides:
cos(x) - 1/2 = (i sqrt(3))/2 or cos(x) - 1/2 = 1/2 (-i) sqrt(3)

Add 1/2 to both sides:
cos(x) = 1/2 + (i sqrt(3))/2 or cos(x) - 1/2 = 1/2 (-i) sqrt(3)

Take the inverse cosine of both sides:
x = cos^(-1)(1/2 + (i sqrt(3))/2) + 2 π n_1 for n_1 element Z or x = 2 π n_2 - cos^(-1)(1/2 + (i sqrt(3))/2) for n_2 element Z
 or cos(x) - 1/2 = 1/2 (-i) sqrt(3)

Add 1/2 to both sides:
x = cos^(-1)(1/2 + (i sqrt(3))/2) + 2 π n_1 for n_1 element Z
 or x = 2 π n_2 - cos^(-1)(1/2 + (i sqrt(3))/2) for n_2 element Z
 or cos(x) = 1/2 - (i sqrt(3))/2

Take the inverse cosine of both sides:
Answer: |
 | x = cos^(-1)(1/2 + (i sqrt(3))/2) + 2 π n_1 for n_1 element Z
 or x = 2 π n_2 - cos^(-1)(1/2 + (i sqrt(3))/2) for n_2 element Z
 or x = cos^(-1)(1/2 - (i sqrt(3))/2) + 2 π n_3 for n_3 element Z or x = 2 π n_4 - cos^(-1)(1/2 - (i sqrt(3))/2) for n_4 element Z

 Jan 7, 2017
 #4
avatar+37091 
+5
Best Answer

Well, I should have specified no REAL solution.....

Yikes...you DO like those long calcs don't you ???

Thanx !!

ElectricPavlov  Jan 7, 2017

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