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hardequalsmath

  • 2 years ago

Find all real numbers: 3sin^2x-sinx=0 HELP! I just basically need help knowing when i add/sub pi or 2pi

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  1. hashim12atd
    • 2 years ago
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    6sin^2 (x) cos^2 (x) -sinx=0 Take sinx common you get sin(x)=0 As you know sin(n pi) =0 so x= n*pi where n is an integer... For remaining portion 6sinx cos^2 (x)-1=0 we get sinx cos^2 x=1/6 as cos^2 x= 1-sin^2 x sinx -sin^3 x=1/6 solve the cubic equation

  2. campbell_st
    • 2 years ago
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    why not just factor out sin(x) \[\sin(x)[3\sin(x) -1] = 0\] so fins the values where sin(x) = 0 and sin(x) = 1/3 they are the solutions.

  3. hardequalsmath
    • 2 years ago
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    i lknow you take the inverse sin for both but then it tells me to sub tract the inverse of 1/3 from pi....and idk why

  4. campbell_st
    • 2 years ago
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    will if you know the sin curve it looks like |dw:1373607480309:dw| so that covers sin(x) = 0 now for the other angle \[x = \sin^{-1}(\frac{1}{3})\] just type it into a calculator.

  5. campbell_st
    • 2 years ago
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    well 1st find your angle then convert it to radians \[1^o = \frac{\pi}{180}... radians\] so you will need to convert your answer. \[answer \times \frac{\pi}{180}..radians\] then when you get you answer, which will be 1st quadrant the other angle will be pi - answer in radians

  6. hardequalsmath
    • 2 years ago
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    how do you know its pi and not 2pi?

  7. campbell_st
    • 2 years ago
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    1st quadrant angle in degrees angle in radians 2nd quadrant angles are 180 - angle or pi - angle 3rd quadrant 180 + angle or pi + angle 4th quadratn 360 - angle or 2pi - angle

  8. hardequalsmath
    • 2 years ago
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    oh wow thanks so much you dont know how long ive been lost....:) thanks

  9. campbell_st
    • 2 years ago
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    glad to help...

  10. hardequalsmath
    • 2 years ago
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    so just to clarify for first quadrant....its just the answer you get from the inverse of sin?

  11. campbell_st
    • 2 years ago
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    thats correct... then 2nd quadrant is 180 - that angle for degrees or pi - angle in radians

  12. hardequalsmath
    • 2 years ago
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    ok thanks

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