You are not logged in.
Pages: 1
I'm in tears over this homework (think the stress is getting to me...)
1)
a) Given that 5sinθ=2cosθ, find the value of tanθ
b) Solve, for 0≤ x<360° 5sin2x=2cos2x
giving your answers 1dp
2)
a) Show that the equation tan2x=5sin2x can be written in the form (1-5 cos2x) sin 2x=0
b) Hence solve, for 0≤x≤180° tan2x=5sin2x to 1dp, showing clearly how you obtained yours answers
Offline
hi Beverrrn
Hey! No tears. Fun is what we're here for.
Looks like no one has told you that
So if 5sinθ=2cosθ divide by cos and also by 5, and you'll have sin/cos = tan = a fraction straight away.
So then part b falls into place fairly easily. Solve for 2x and thus for x. Make sure you find all the answers.
Q2. Re-write as sin(2x) / cos(2x) - 5sin(2x) = 0 and then is should be possible for you to finish.
If you post your answers, I'll check them for you.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Pages: 1