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anonymous
 one year ago
find all radian solutions in the interval 0<= x < 2pi round to four decimal places.
4sin^2x12sinx1=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?
anonymous
 one year ago
find all radian solutions in the interval 0<= x < 2pi round to four decimal places. 4sin^2x12sinx1=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?

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[4\sin^2(x)12 \sin(x)1=0 \\ \sin(x)=\frac{(12) \pm \sqrt{(12)^24(4)(1)}}{2(4)} \\ \sin(x)=\frac{12 \pm \sqrt{144+16}}{8} \\ \sin(x) =\frac{12 \pm \sqrt{160}} {8} \\ \sin(x)=\frac{12 \pm 4 \sqrt{10}}{8} \\ \sin(x)=\frac{3 \pm \sqrt{10}}{2} \] 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1however you should see (3+sqrt(10))/2 >1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so you only have to solve \[\sin(x)=\frac{3  \sqrt{10}}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! That helped me out alot!
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