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Posted by on Wednesday, March 23, 2011 at 10:22pm.

tan(x)=5 sin(x) for interval -π < x < π

A) 0, 1.571 B) -1.571, 0, 1.571 C) -1.369, 0, 1.369 D) 0, 1.369

  • Pre Calc - , Wednesday, March 23, 2011 at 10:37pm

    recall that tan(x) can be rewritten as
    tan (x) = sin (x) / cos (x)
    substituting:
    sin(x) / cos(x) = 5 sin(x)
    the sin(x) will be cancelled:
    1/cos(x) = 5
    cos(x) = 1/5
    solving this,
    x = +/- 1.369
    since it must be on interval -π < x < π
    x = - 1.369

  • Pre Calc - , Wednesday, March 23, 2011 at 10:39pm

    are we solving ????

    tanx = 5sinx
    sinx/cosx= 5sinx
    sinx = 5sinxcosx
    sinx - 5sinxcos)=0
    sinx(1 - 5cosx) = 0
    sinx = 0 or cosx = 1/5

    if sinx = 0, x = 0, π or 2π

    if cosx = 1/5, x = 1.369 or -1.369 if -π < x < π

    so for the given domain
    x = -1.369 , 0, 1.369 , which would be choice C)

  • Pre Calc - , Wednesday, March 23, 2011 at 10:52pm

    Thanks so much guys!!!!

  • Pre Calc - , Wednesday, March 23, 2011 at 10:58pm

    Did you notice that Jai missed one of the answers of
    x = 0.
    You should not cancel sinx , but rather use it as one of the factors.
    by canceling sinx , he "lost" the answer to sinx = 0

  • Pre Calc - , Wednesday, March 23, 2011 at 11:18pm

    oh yeah,, sorry about that. 0 is also a solution~
    thanks for correcting me, sir~ :)

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