6sin^2(x) - 5sin(x) + 1 = 0
solve
6 answers
*** 6sin^2(x) + 5sin(x) + 1 = 0
It factors
(2sin x + 1)(3sin x + 1) = 0
sinx = -1/2 or sinx = -1/3
If sinx = -1/2, I know x must be in III or IV
x = 210° or 330°
x = 7π/6 or 11π/6
if sinx = -1/3, x must be in III or IV
angle in standard position is appr 19.47°
x = 199.47° or 340.53°
x = 3.48 radians or 5.94 radians
(2sin x + 1)(3sin x + 1) = 0
sinx = -1/2 or sinx = -1/3
If sinx = -1/2, I know x must be in III or IV
x = 210° or 330°
x = 7π/6 or 11π/6
if sinx = -1/3, x must be in III or IV
angle in standard position is appr 19.47°
x = 199.47° or 340.53°
x = 3.48 radians or 5.94 radians
but how did you get x = 7π/6 or 11π/6 when x=sin-1 (-1/2) is -π/6
normally our domain is 0 ≤ x ≤ 2π
perhaps if you think in degrees it might be easier to see.
-π/6 radians is -30°
now where is -30° ?
doesn't it coincide with 330°,
and isn't 330° = 11π/6 radians ?
perhaps if you think in degrees it might be easier to see.
-π/6 radians is -30°
now where is -30° ?
doesn't it coincide with 330°,
and isn't 330° = 11π/6 radians ?
so then how did you derive that just by knowing?
what im trying to say is how would i be able to come about that in a question that was similar like this?
what im trying to say is how would i be able to come about that in a question that was similar like this?
Let's consider how I got the values of x
we started with sinx = -1/2
We know from our basic trig functions or by the CAST rule, that x must be in quadrants III or IV
(you must know that!)
We then find the angle in standard position, that is the corresponding angle if it had been in the first quadrant, so find
sin^-1 (+1/2) to get x = π/6 or 30°
so our first angle is going to be in quad III
which is π + π/6 = 7π/6
our second angle is 2π - π/6 = 11π/6
of course there will an infinite number of solutions, and the angle -π/6 that your calculator gives you is just one of those, but it is outside of our 0 ≤ x ≤ 2π domain.
Your calculator has been programmed to give you one angle, and it will be the angle closest to zero.
we started with sinx = -1/2
We know from our basic trig functions or by the CAST rule, that x must be in quadrants III or IV
(you must know that!)
We then find the angle in standard position, that is the corresponding angle if it had been in the first quadrant, so find
sin^-1 (+1/2) to get x = π/6 or 30°
so our first angle is going to be in quad III
which is π + π/6 = 7π/6
our second angle is 2π - π/6 = 11π/6
of course there will an infinite number of solutions, and the angle -π/6 that your calculator gives you is just one of those, but it is outside of our 0 ≤ x ≤ 2π domain.
Your calculator has been programmed to give you one angle, and it will be the angle closest to zero.
1 answer