Posted by Abby 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 - Jai, Wednesday, March 23, 2011 at 10:37pm
recall that tan(x) can be rewritten as
tan (x) = sin (x) / cos (x)
sin(x) / cos(x) = 5 sin(x)
the sin(x) will be cancelled:
1/cos(x) = 5
cos(x) = 1/5
x = +/- 1.369
since it must be on interval -π < x < π
x = - 1.369
- Pre Calc - Reiny, Wednesday, March 23, 2011 at 10:39pm
are we solving ????
tanx = 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 - Abby, Wednesday, March 23, 2011 at 10:52pm
Thanks so much guys!!!!
- Pre Calc - Reiny, 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 - Jai, 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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