Question

Trigonometric Equation

2sin^2A + cosA + 1 = 0

Answer 1

use the relation,
\(\Large \sin^2A+\cos^2A=1\)
to get a quadratic equation in only cos A

Answer 2

How's that @hartnn
I really dont know

Answer 3

2sin^2A + cosA + 1 = 0

2 (1- cos^2 A) +cos A +1 =0

did u get this?

Answer 4

Yes. Got that part
Then what?
@hartnn

Answer 5

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 ?

Answer 6

Okay. Will it be

-2x^2+x+3

?
@hartnn

Answer 7

yes,
or you can write that as
2x^2-x-3=0 too
can you solve this quadratic?

Answer 8

I dont know for cosA=3/2
???

But for cos A=-1
A={180}
It it correct?
@hartnn

Answer 9

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

Answer 10

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 :)

Answer 11

else, the general form will be A = 180 ± 360n

What does that mean??? ^
So A is just 180 nothing else?
@hartnn

Answer 12

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 :) )

Answer 13

Okay. I see. Thank you

Answer 14

welcome ^_^