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anonymous

  • one year ago

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?

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  1. freckles
    • one year ago
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    \[4\sin^2(x)-12 \sin(x)-1=0 \\ \sin(x)=\frac{-(-12) \pm \sqrt{(-12)^2-4(4)(-1)}}{2(4)} \\ \sin(x)=\frac{12 \pm \sqrt{144+16}}{8} \\ \sin(x) =\frac{12 \pm \sqrt{160}} {8} \\ \sin(x)=\frac{12 \pm 4 \sqrt{10}}{8} \\ \sin(x)=\frac{3 \pm \sqrt{10}}{2} \] 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

  2. freckles
    • one year ago
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    however you should see (3+sqrt(10))/2 >1

  3. freckles
    • one year ago
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    so you only have to solve \[\sin(x)=\frac{3 - \sqrt{10}}{2}\]

  4. anonymous
    • one year ago
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    Thank you so much! That helped me out alot!

  5. freckles
    • one year ago
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    np

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