anyone know this one?

[Anonymous No.2238239]

knowing that 1 is root to the equation below
cos²(a)*x² - (4*cos(a)*sin(b))*x + (3/2)sin(b) = 0

what could be 'a' and 'b'?
a) pi/8 and 3*pi/8
b) pi/6 and pi/3
c) pi/4 and pi/4
d) pi/3 and pi/6
e) 3*pi/8 and pi/8

[Barack !446r2OBAMA No.2238247]

Learn to Latex, bro.

[Anonymous No.2238273]

Latex-ing for OP
<div class="math">cos^2 (a)x^2 - 4xcos (a)sin(b) + \frac{3}{2}sin(b)</div>Shouldn't it be sin^2 at the end though?

[Anonymous No.2238286]

>>2238273
no, i just checked it...

[Anonymous No.2238292]

op here
the book says the answer is D, but i have no idea how to get there

[Anonymous No.2238325]

>>2238292
There's probably trig identities you could fuck around with until you get a more simple equation.

when you say 1 is a root, does that mean a possibly value for x is 1?

[Anonymous No.2238336]

>>2238325
i think so

[Anonymous No.2238417]

>>2238247
teach me how to latex.
where to learn it? links?

[Anonymous No.2238421]

>>2238417
http://www.codecogs.com/latex/eqneditor.php

play around, you don't need a tutorial

[Anonymous No.2238440]

Remainder after dividing cos^2(a)*x^2 - (4*cos(a)*sin(b))*x + (3/2)sin(b) by (x-1) is

-4cos(a) sin(b) + cos(a)^2+3sin(b)/2

plug in the options to see which make this zero.