How can I solve for b in the equation csc(6b+pi/8) = sec(2b-pi/8)?

[I Like Pi]

Solving trig function...

Homework Statement

given, $$csc(6b+pi/8) = sec(2b-pi/8)$$

solve for b

Homework Equations

The Attempt at a Solution

I managed to simplify it to:

$$cos(2b-pi/8) = sin(6b+pi/8)$$

How would i solve for b

well i know that $$sinx = cos(pi/2-x)$$ so...
$$cos(2b-pi/8) = cos[pi/2-(6b+pi/8)]$$
$$cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0$$
$$cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0$$
$$8b-pi/2 = cos-1(0)$$
$$8b = pi$$
$$b=pi/8$$

but it doesn't work :sad:

 

[Mentallic]

$$cos^{-1}(cosA+cosB)\neq A+B$$

which is what you have done.

And just as a word of advice, if you do happen to do something to the equation such as take the inverse cosine of it, you need to do it all in one go - i.e. don't take the inverse of one side, then the inverse of the other side. Do it altogether.

Using $$sin(x)=cos(\pi/2-x)$$

$$sin(6b+\pi/8)=cos(3\pi/8-6b)=cos(-3(2b-\pi/8))$$

Now use the fact that $$cos(x)=cos(-x)$$ and you should be all good from there.