Question

find all radian solutions in the interval 0<= x < 2pi round to four decimal places.
4sin^2x-12sinx-1=0 i did the quadratic formula and got 12 +- square root of 160, which breaks down into 12 +- 4 square root 10. im not sure how im supposed to get that into radians, or how ro round it 4 decimal places. am i doing this right?

Answer 1

\[4\sin^2(x)-12 \sin(x)-1=0 \\ \sin(x)=\frac{-(-12) \pm \sqrt{(-12)^2-4(4)(-1)}}{2(4)} \\ \sin(x)=\frac{12 \pm \sqrt{144+16}}{8} \\ \sin(x) =\frac{12 \pm \sqrt{160}} {8} \\ \sin(x)=\frac{12 \pm 4 \sqrt{10}}{8} \\ \sin(x)=\frac{3 \pm \sqrt{10}}{2} \]
ok just wanted to check your answer so far
now assume a is between -1 and 1
and sin(x)=a
then
\[x=\arcsin(a)+2n \pi \text{ or } x=-\arcsin(a)+(2n+1)\pi\]
where n is an integer

Answer 2

however you should see (3+sqrt(10))/2 >1

Answer 3

so you only have to solve
\[\sin(x)=\frac{3 - \sqrt{10}}{2}\]

Answer 4

Thank you so much! That helped me out alot!

Answer 5

np