Posted by Suzy on Friday, August 16, 2013 at 9:36pm.
2. solve cos 2x3sin x cos 2x=0 for the principal values to two decimal places.
3. solve tan^2 + tan x1= 0 for the principal values to two decimal places.
4. Prove that tan^2(x) 1 + cos^2(x) = tan^2(x) sin^2 (x).
5.Prove that tan(x) sin(x) + cos(x)= sec(x)
6.Prove that tan(x) cos^2(x)+sin^2(x)= cos(x)+ sin(x)
7.Prove that 1+tan(x)/1tan(x)= sec^2(x)+ 2tan(x)/1tan^2(x)
8.Prove that sin^2(x)cos^2(x)/tan(x)sin(x)+cos(x)tan(x)=cos(x)con(x)cos(x)
9. find a counterexample to show that the equation sec(x)cos(x)=sin(x) sec(x) is not an identity

Trigonometry desperate help, clueless girl here  Steve, Saturday, August 17, 2013 at 12:37am
#2
cos2x  3sinx cos2x = 0
(cos2x)(13sinx) = 0
as you know, if the product of two numbers is zero, one or the other must be zero. So, cos2x = 0 or 13sinx = 0
cos2x=0 means x is pi/4,3pi/4,5pi/4,7pi/4
13sinx=0 means x = arcsin(1/3) = .3398
But, you need all angles between 0 and 2pi, so since sinx >0 in Qi and QII,
x = .3398 or pi.3398=2.8018
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#3.
did this one already also. What was unclear?
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#4
possibly the most useful trig identity is sin^2 x + cos^2 x = 1. You have
tan^2 x  1 + cos^2 x
tan^2 x  (1cos^2 x)
tan^2 x  sin^2 x
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#5.
tanx sinx + cosx
sinx/cosx * sinx + cosx
(sin^2 x + cos^2 x)/cosx
1/cosx
secx
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#6.
Must be a typo. If x=pi/4,
tan(x) cos^2(x)+sin^2(x) = 1*1/2 + 1/2 = 1
cos(x)+sin(x) = 1/√2+1/√2 = √2
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#7.
(1+tanx)/(1tanx)
(1+tanx)^2 / (1tan^2 x)
(1 + 2tanx + tan^2 x)/(1tan^2 x)
(sec^2 x + 2tanx)/(1tan^2 x)
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#8.
(sin^2 x  cos^2 x)/(tanx*sinx + cosx*tanx)
(sinxcosx)(sinx+cosx)/(tanx(sinx+cosx))
(sinxcosx)/tanx
sinx*cotx  cosx*cotx
cosx  cosx*cotx
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#9.
Usually a familiar angle will do the trick. If x=pi/4,
secxcosx = √2  1/√2
sinx*secx = 1/√2*√2 = 1
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