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)

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 -
**Reiny**, 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 -
**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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