
Trig help.
I need help on these questions please.
1. given that sin^2 x + cos^2 x = 1 show that 1+cot^2 x = cosec^2 x
2. Solve for values x in the interval pi<x<pi for the following in 2 decimal places:
a. 2cosec(x/2)  5 =0
b. 2cot^2 x  cot x 7 = 0
Thanks

For the first question I got:
sin^2 x + cos^2 x = 1
so (sin^2 x)/(sin^2 x) + (cos^2 x)/(cos^2 x) = 1/(sin^2 x)
so 1 + (cos x/ sin x)^2 = (1/ sin x)^2
so 1 + cot^2 x = cosec ^2 x.
Is that right?
But im really stuck on the other two questions

1) Divide by $\displaystyle \sin^{2}(x)$ and see what happens.
2a) sin(x) = 1/csc(x)
2b) Solve this first: 2Y^2  Y  7 = 0. I see four solutions in your future.

for 2a I got:
2cosec(x/2)  5 = 0
so 2(1/sin(x/2))  5 = 0
so 2(2/sin x)  5 = 0
so 4/sin x 5 = 0
so 4 = 5sin x
so sin x = 4/5
so x = inverse sin(4/5)
So the only solution I got for x is 53.13 to 2 decimal places.
Is this right?

sorry tk hunny but 2b does not factorise.

I wasn't suggesting factoring, unless that happens to work.
Solve it, nonetheless.