August 22nd, 2018, 10:53 AM  #1 
Member Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2  Trigonometry identities
I need detailed solutions for the equation tan^2(x+45)=5 for 0<x<360 where 45 is in degrees. Last edited by skipjack; August 22nd, 2018 at 11:46 AM. 
August 22nd, 2018, 11:45 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,293 Thanks: 1968 
As tan(x + 45$^\circ$) = ±√5, you can calculate possible values for x + 45$^\circ$, namely ±65.9$^\circ$ approximately. What progress can you make from there? 
August 22nd, 2018, 12:45 PM  #3 
Member Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 
I don't understand please make it simpler&understandable

August 22nd, 2018, 06:37 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,293 Thanks: 1968 
tan²(x+45$^\circ$) = 5 implies tan(x+45$^\circ$) = ±√5. Do you understand that? By use of a calculator, tan$^{1}$(±√5) = ±65.9$^\circ$ approximately. Do you also understand that? Are you aware that tan(A) = tan(A ± k(180$^\circ$)) for any integer value of k and any value of A for which tan(A) is defined? 
August 22nd, 2018, 10:30 PM  #5 
Member Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 
YEAH...proceed please

August 23rd, 2018, 06:13 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,293 Thanks: 1968 
You need values of x + 45$^\circ\!$ between 45$^\circ\!$ and 405$^\circ\!$, so choose all integer values of k such that ±65.9$^\circ\!$ + k(180$^\circ\!$) is within that range, then you can get the desired (approximate) answers by subtraction of 45$^\circ\!$. For example, using 65.9 and k = 1 would lead to (65.9 + 180  45)$^\circ\!$, which is 69.1$^\circ\!$. 

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