+0

# how to solve

0
591
2

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

May 27, 2015

#2
+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....

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
+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
+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....

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