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October 16th, 2017, 08:19 AM  #1 
Newbie Joined: Oct 2017 From: england Posts: 1 Thanks: 0  Trig identities and equations
Any help would be appreciated 2cosecθ + 4.1 = 0 14secθ  2 = 12tan^2 θ Last edited by strugglingalong; October 16th, 2017 at 09:12 AM. 
October 16th, 2017, 09:04 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,774 Thanks: 1428 
you may want to type out the equations. as posted, your image is too small to read. 
October 16th, 2017, 10:24 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
In the second equation, after dividing by 2, replace tan²θ with sec²θ  1 to obtain a quadratic in sec θ.

October 19th, 2017, 05:33 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
cosec(θ)= 1/sin(θ) so the first equation is 2/sin(θ)+ 4.1= 0. Subtract 2/sin(θ) from both sides: 4.1= 2/sin(θ) Multiply both sides by sin(θ): 4.1 sin(θ)= 2 Divide both sides by 4.1: sin(θ)= 2/4.1= 0.48780... θ= arcsin(0.48780...) sec(θ)= 1/cos(θ) and tan(θ)= sin(θ)/cos(θ) so the second equation is 14/cos(θ) 2= 12sin^2(θ)/cos^2(θ). Multiply both sides by cos^2(θ): 14cos(θ) 2cos^2(θ)= 12sin^2(θ) Here, I would use the fact that sin^2(θ)= 1 cos^2(θ) to write that as 14cos(θ) 2cos^2(θ)= 12 12cos^2(θ). Add 12 cos^2(θ) to both sides: 10cos^2(θ)+ 14cos(θ)= 12. Subtract 12 from both sides: 10cos^2(θ)+ 14cos(θ) 12= 0. That's a quadratic equation in cos(θ). Just to make it easier to read, let x= cos(θ). Then the equation becomes 10x^2+ 14x 12= 0. Divide by 2: 5x^2+ 7x 6= 0. Solve that quadratic equation using, say, the quadratic formula: x= (7+/ sqrt(49 (4)(5)(6)))/10= (7+/ sqrt(49+ 120))/10= (7+/ sqrt(169))/10= (7+/ 13)/10. Taking the "+", x= (7+ 13)/10= 6/10= 0.6 Taking the "", x= (7 13)/10= 20/10= 2. Of course the cosine of any real number cannot be less than 1 so the only valid answer is x= cos(θ)= 0.6. θ= arccos(0.6).  

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