*Trigonometry Equations*

• Oct 22nd 2011, 12:02 PM
JohnTerry
*Trigonometry Equations*
Hey guys!

Could you explain how to resolve, step by step, the following equations(or at least lines d) and e) )

http://www.mathhelpforum.com/math-he...022_210239.jpg

• Oct 22nd 2011, 12:10 PM
SammyS
Re: *Trigonometry Equations*
Show us some work.

What do you have?

Where are you stuck?
• Oct 22nd 2011, 12:31 PM
JohnTerry
Re: *Trigonometry Equations*
Quote:

Originally Posted by SammyS
Show us some work.

What do you have?

Where are you stuck?

sen(x+x/4) = -1/2

5x/4 = -π/6 + 2πK V 7π/6 +2πk, KeZ

x= -4π/30 + 8/5πk V 28π/30 + 8/5πk, keZ

x= -2/15π +8/5π V 14/15π + 8/5πk, keZ

e)
senxcosx = senx/2
• Oct 22nd 2011, 12:37 PM
e^(i*pi)
Re: *Trigonometry Equations*
Quote:

Originally Posted by JohnTerry

e)
senxcosx = senx/2

Assuming sen(x) = sin(x)

It is better to factorise so you do not inadvertently divide by 0 or discard solutions.

$\sin(x)- 2 \sin(x)\cos(x) =0 \Leftrightarrow \sin(x)(2-\cos(x))=0$
• Oct 22nd 2011, 12:51 PM
JohnTerry
Re: *Trigonometry Equations*
Quote:

Originally Posted by e^(i*pi)
Assuming sen(x) = sin(x)

It is better to factorise so you do not inadvertently divide by 0 or discard solutions.

$\sin(x)- 2 \sin(x)\cos(x) =0 \Leftrightarrow \sin(x)(2-\cos(x))=0$

But where did the first sin(x) go?

It is possible to solve the equation if you factorise it, but then we assume that the first sin(x)=0? Then, sin(x)=0; sin(x)=0 ; cos(x)=2

And can you please confirm the first equation(d). The answer in the book is different.

• Oct 22nd 2011, 01:03 PM
mr fantastic
Re: *Trigonometry Equations*
Quote:

Originally Posted by JohnTerry
Hey guys!

Could you explain how to resolve, step by step, the following equations(or at least lines d) and e) )