+0  
 
0
591
2
avatar

tan x + 4 sec x = 4 , 0<x<360

 May 27, 2015

Best Answer 

 #2
avatar+94619 
+5

 

 

 

Ah....Mathcad......you did very well but you gave up a little too quickly, here........!!!!......notice the following factorization....

 

15sec^2(x)-32sec(x)+17=0

 

(15sec(x) - 17) (sec(x) -1)  = 0

 

So either......

 

sec(x) = 1  which happens at 0 and 360.....but, both of these are out of the requested interval....or....

 

sec(x)  = 17/15   which happens at about 28.072°  and about 331.928°

 

Here's a graph..........https://www.desmos.com/calculator/fpewbunwdj

 

Notice that the solution of  28.072°  is "extraneous"......this will frequently happen when we square both sides of an equation......the only  "good" solution in the requested interval occurs at about 331.928°

 

 

 

 May 28, 2015
 #1
avatar+122 
+5

tan(x)+4sec(x)=4

tan(x)=4(1-sec(x))   Now square both sides to get

tan^2(x)=16(1-sec(x))^2  and using the identity tan^2(x)=sec^2(x)-1 on left hand side

sec^2(x)-1=16(1-sec(x))^2   Expanding brackets and collecting terms gives

15sec^2(x)-32sec(x)+17=0.

Now just solve the quadratic in sec(x) to get the 2 values of the solution. Solution has real roots as b^2-4ac=1024-1020=4,but formula will need to be used as this will not factorise. I am now going for beer.

 May 27, 2015
 #2
avatar+94619 
+5
Best Answer

 

 

 

Ah....Mathcad......you did very well but you gave up a little too quickly, here........!!!!......notice the following factorization....

 

15sec^2(x)-32sec(x)+17=0

 

(15sec(x) - 17) (sec(x) -1)  = 0

 

So either......

 

sec(x) = 1  which happens at 0 and 360.....but, both of these are out of the requested interval....or....

 

sec(x)  = 17/15   which happens at about 28.072°  and about 331.928°

 

Here's a graph..........https://www.desmos.com/calculator/fpewbunwdj

 

Notice that the solution of  28.072°  is "extraneous"......this will frequently happen when we square both sides of an equation......the only  "good" solution in the requested interval occurs at about 331.928°

 

 

 

CPhill May 28, 2015

14 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.