Maths HW Problem

[xChrisVx]

Hi

I'm pretty stuck on this maths homework problem which is due in for tuesday

Here is the question

Factorise the following expressions and solve to find x : 0° ≤x ≤360°

(a) 2〖sin〗^(2 ) x-sinx-3=0

(b) 2〖cos〗^2 x-5cosx+2=0

any ideas are greatly appeciated

[Sonic Boom]

a) We must factorize this crap first, so let sinx = x.

So we have 2x^2 - x - 3 = 0
This the same as (2x - 3)(x+1) = 0

This equation is true if x = 3/2 or if x = -1

So x = 3/2 or -1, but sin(x) = x. So Sin(x) = 1.5 or -1

Use your calculator or a sin(x) graph to find the final answer in degrees. In your question it says x can only be between 0 and 360 degrees.

b) Same procedure here.

2x^2 - 5x + 2 = 0

This is the same as (2x - 1)(x -2) = 0 if you multiply it out.

Hence, x = 1/2 or x = 2.......you know what to do next.......

Hope that helps brazzo.

[xChrisVx]

Fantastic!!!!! Thanks very much!!!!!!!! This has at least got me started.

However Sin-1 (1.5) isn't possible and Sin-1 (-1) = -90 which isn't in the range of the question

Sorry, that's me being stupid. I never realised that Sin-1 (-1) = 270

Thanks for the help

[FullyAlive]

You've done it now but for future reference when solving trig identities such as sine, cosine and tangent expressions you can use the graph to look for the other answers for example 30° on a sine graph if you look at it you'll see that the curve hits that point at 150° too. With practice it's relatively easy to learn the patterns of each graph.