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I am so stuck with maths c2 trig

I would like help with 1 trig identity question. I can't get it to fit?
sdfsdfsdf.JPG5B and 6B
Reply 2
Avatar for user786
user786
Original post by ineedhelpwithph
sdfsdfsdf.JPG5B and 6B


Use substitution for both. 5B is in the firm of a quadratic equation.
Original post by user786
Use substitution for both. 5B is in the firm of a quadratic equation.


Still confused? :frown:
Reply 4
Avatar for dnr_23
dnr_23
9
split the fraction into two different ones over 1-sin^2(x)
Reply 5
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dnr_23
9
Original post by dnr_23
split the fraction into two different ones over 1-sin^2(x)


just realised you wanted 5b haha my bad
Reply 6
Avatar for dnr_23
dnr_23
9
for 5b)
equate tan^2(x)-1 to 5-tan(x). your quadratic will factorise so you will get tan(x) = a and tan(x) = b
then do the inverse tan for a and then b
and then do 180+the angle that you get for each root
(edited 7 years ago)
Reply 7
Avatar for Naruke
Naruke
11
Original post by ineedhelpwithph
Still confused? :frown:


5B) Since you found out that sin⁡2x−cos⁡2x1−sin⁡2x=tan⁡2x−1\frac{\sin^2x - \cos^2x}{1-\sin^2x} = \tan^2x -1, you can sub it into your equation tan⁡2x−1=5−tan⁡x \tan^2x - 1 = 5 - \tan x . Now all you need to do is rearrange and you will get a quadratic equation in terms of tan⁡x \tan x . If solving that seems daunting, let y=tan⁡x y = \tan x
Reply 8
Avatar for dnr_23
dnr_23
9
6c)
factorise your quadratic in terms of cos(x) your roots will be -2 and 1/3
inverse cos to find the angles. you will get math error for cos^-1(-2) because the cos curve has a minimum at -1.
your angles for x will therefore be in the 1st and 4th quadrants as you will need to do 360-ANS.

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