Question

solve algebraically 2sin^2 theta-4sin theta=cos^2 theta-2

Answer 1

replace cos^2 = 1-sin^2
let everything in LHS
solve quadratic polynomial with sin theta is variable

Answer 2

hey, I guide only, I don't work for you. You do it,

Answer 3

what you think im doing now

Answer 4

THANKS REALLY HELPED ALOT

Answer 5

glad to hear that THANKS REALLY HELPED ALOT, LOL, LOL LOL.... , actually what does it mean? sarcastic? why do you shout at me with your appreciation? heheheha

Answer 6

I SHOUT WITH APPRECIATION BECAUSE I FINALLY UNDERSTAND IT!!!

Answer 7

hahahahaa......

Answer 8

trig is hard so if i understand one thing im excited

Answer 9

sometimes, trig is trick, too

Answer 10

yeah sometimes i just want to burn my trig notebook

Answer 11

2sin^2 theta-4sin theta=cos^2 theta-2
sin^2(x) + sin^2(x) - cos^2(x) +2- 4sin(x) = 0
sin^2(x) + cos^2(x) + sin^2(x) + cos^2(x) - 3cos^2(x) + 2 - 4sin(x) =0
sin^2(x) + cos^2(x) =1
-3[cos^2(x) + (4/3)sin(x) - 4/3]
-3[1 - sin^2(x) +(4/3)sin(x) - 4/3] = 3[sin^2(x) - (4/3)sin(x) + 1/3] = 0
set y = sin(x) solve quadratic y^2 - (4/3)y -1/3 = (y - 1/3)(y -1) = 0
sin(x) = y = 1, 1/3
therefore x = pi/2 = 90 deg, 19.47 deg
Did anyone else get these answers?