trig equation help

Tweety · #1 $Old$ November 2nd, 2009, 11:38 AM

Solve in the interval $0\leq x \leq 2\pi$ ,

$cosec(x + \frac{\pi}{15} ) = -\sqrt{2}$

not really sure what to do,

I expanded the brackets and got

$cosecx + cosec\frac{\pi}{15} = -\sqrt{2}$

anything I try from here just gives me very wrong answers, can anyone suggest what I do next?

thank you.

e^(i*pi) · #2 $Old$ November 2nd, 2009, 11:46 AM

Quote:

Originally Posted by Tweety $View Post$

Solve in the interval $0\leq x \leq 2\pi$ ,

$cosec(x + \frac{\pi}{15} ) = -\sqrt{2}$

not really sure what to do,

I expanded the brackets and got

$cosecx + cosec\frac{\pi}{15} = -\sqrt{2}$

anything I try from here just gives me very wrong answers, can anyone suggest what I do next?

thank you.

You don't expand trig functions like that.

$\frac{1}{sin\left(x+\frac{\pi}{15}\right)} = -\sqrt{2}$

Take the reciprocal.

$sin\left(x+\frac{\pi}{15}\right) = -\frac{1}{\sqrt2}$

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