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October 16th, 2017, 08:19 AM   #1
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Trig identities and equations

Any help would be appreciated

2cosecθ + 4.1 = 0
14sec⁡θ - 2 = 12tan^2 θ
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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 tan²⁡θ with sec²⁡θ - 1 to obtain a quadratic in sec θ.
October 19th, 2017, 05:33 AM   #4
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Quote:
 Originally Posted by strugglingalong Any help would be appreciated 2cosecθ + 4.1 = 0 14sec⁡θ - 2 = 12tan^2 θ
My first reaction with problems like these is to replace everything with sine and cosine:

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.

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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