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kaylala
Trigonometric Equation 2sin^2A + cosA + 1 = 0
use the relation, \(\Large \sin^2A+\cos^2A=1\) to get a quadratic equation in only cos A
How's that @hartnn I really dont know
2sin^2A + cosA + 1 = 0 2 (1- cos^2 A) +cos A +1 =0 did u get this?
Yes. Got that part Then what? @hartnn
right, now thats an quadratic equation in cos A. imagine you put cos A = x 2(1-x^2)+x+1 =0 can you bring that in the form of ax^2+bx+c=0 ?
Okay. Will it be -2x^2+x+3 ? @hartnn
yes, or you can write that as 2x^2-x-3=0 too can you solve this quadratic?
I dont know for cosA=3/2 ??? But for cos A=-1 A={180} It it correct? @hartnn
if you need solutions in the range 0 to 360, then yes, A = 180 is correct. else, the general form will be A = 180 \(\pm\) 360n
as for cos A = 3/2 recollect that cos A can never be greater than 1 as the range of cos function is -1 to 1 so, cos A = 3/2 will have no solution :)
else, the general form will be A = 180 ± 360n What does that mean??? ^ So A is just 180 nothing else? @hartnn
with cos A = -1 you got A = 180, right ? what about when A = 180+360 = 540 ? is cos 540 also = -1 ? if yes, why don't you consider it as a solution ? similarly for 180-360,180+2*360, 180-2*360 and so on (we don't consider them as the solution only when it is given that the range of A is from 0 to 360....and in that case only 180 is your answer :) )