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Please help me with trig. Please explain how you got the answer so i can learn.?

Find all exact solutions that exist on the interval [0, 2π).

tan(x) sin(x) + sin(−x) = 0

3 Answers

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  • alex
    Lv 7
    2 months ago

    Hint:

    sin(-x)=-sinx

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  • 2 months ago

    Answer is  x = 0, π/4, π , 5π/4 , 2π 

    See how --

    tan(x) sin(x) + sin(−x) = 0

    We know that sin ( - x )  =  - sin x

    tan(x) sin(x) + sin(−x) = 0

    => tan(x) sin(x) - sin(x)  = 0

    => ( sin x/cos x ) * ( sin x ) - sin x   =  0

    => ( sin x ) ( sin x/cos x  - 1 )   =   0

    => ( sin x ) ( tan x - 1 )  =  0

    => ( sin x )   =  0   Or  tan x  =  1

    For sin x  =  0  ...... x  =  0 , π , 2π

    For  tan x  = +1 ....... x  =  π/4 , 5π/4

    Hence  x  =  0, π/4, π , 5π/4 , 2π  .............. Answer

  • rotchm
    Lv 7
    2 months ago

    You (should) know that sin(-x) = -sin(x). Know this BY HEART. 

    Thus, your eqs becomes

    tan(x)sin(x) - sin(x) = 0.  Now, factor out sin(x) to get

    sin(x) * (tan(x) - 1) = 0

    Thus sin(x) = 0 and tan(x) - 1 = 0.

    Can you solve each of these now?

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