# Trigonometric graph question

1. Apr 13, 2007

### Trail_Builder

just encountered this question and kinda confused at how to solve it since I havn't bin told and havn't worked it out for myself. hope you can help.

1. The problem statement, all variables and given/known data

Use the graphs (shows 2 graphs) to find the values of x in the range $$0 /leq x /leq 720$$ when 2sinx = cosx -1

2. Relevant equations

3. The attempt at a solution

I found from the graph that x could equal 0, 360, and 720. but the 2 lines cross at another point before x=360 so there should be 2 more values of x to satisfy the question.

however, i can't easily read off an accurate result. i am only 16 so do you rekon they allow for error reading off the graph instead of working it out a very accurate way?

thnx

2. Apr 13, 2007

### Gib Z

3. Apr 13, 2007

### HallsofIvy

Graphing calculators will allow you to "zoom" in on a point so if you have one that should give you very accurate solutions. It is, however, possible to get "exact" solutions. $cos x= \sqrt{1- sin^2 x}$ so you can rewrite the equation as $2 sin x= \sqrt{1- sin^2 x}- 1$. Now, just to simplify the writing, let y= sin x. The equation is $2 y= \sqrt{1- y^2}- 1$. Add 1 to both sides: $2y+ 1= \sqrt{1- y^2}$ and, finally, squaring, $4y^2+ 4y+ 1= 1- y^2$ or $5y^2+ 4y= y(5y+4)= 0$. One solution to that is y= sin x= 0 (and we must also have cos x= 2sin x+ 1= 1). that gives you multiples of 360 as solutions. If y is not 0 then we must have 5y+ 4= 0 so y= sin x= -4/5 (and cos x= 2sin x+ 1= -8/5+1= 3/5. x= Arcsin(-4/5)+ multiples of 360.