# Thread: Simple trig equation confusion

1. ## Simple trig equation confusion

Hi, I'm a little confused over the following question:

3sin2x-1=0 on the interval XE[0,2pi]

Working it out, I got:

2x=sin-1(1/3)

However, the answers show that there are four solutions: 0.17, 1.4, 3.31, 4.54

Most of the previous questions have only had 2 solutions and the only question I've encountered with more included cubes.

Could anyone help me figure out where the other two solutions came from and what I should be looking for in future questions of the same nature?

Thanks!

2. Hello, youngb11!

$3\sin2x-1\:=\:0\:\text{on the interval }x \in [0,\,2\pi]$

First, look at the interval: . $[0,\,2\pi]$

We want $\,x$ to be between 0 and 6.28

We have: . $\sin2x \:=\:\frac{1}{3} \quad\Rightarrow\quad 2x \:=\:\sin^{\text{-}1}\!\!\left(\frac{1}{3}\right)$

Recall that the inverse sine has an infinite number of values.

$\text{So we have: }\;2x \;=\;\begin{Bmatrix}\text{-}3.48\; \\ 0.34 \\ 2.80 \\ 6.62 \\ 9.08 \\ 12.90 \end{Bmatrix}$

$\text{Divide by 2: }\;x \;=\;\begin{Bmatrix}\text{-}1.74\; \\ 0.17 \\ 1.14 \\ 3.31 \\ 4.54 \\ 6.45 \end{Bmatrix}$

We want the values between 0 and 6.28, remember?

$\text{Therefore: }\;x \;=\;\begin{Bmatrix}0.17 \\ 1.14 \\ 3.31 \\ 4.54 \end{Bmatrix}$

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It's the "2" in sin 2x that causes the trouble.

$\text{Example: \:Solve }\sin3x \:=\:\frac{1}{2}\;\text{ for }x \in [0^o,\;360^o]$

$\text{We know that: }\:3x \:=\:\sin^{\text{-}1}\left(\frac{1}{2}\right) \;=\;30^o,\:150^o$

But because of the "3" in 3x, we must "go around 3 times": .**

. . $3x \;=\;\begin{Bmatrix} 30^o \\ 150^o \\ \\[-3mm]390^o \\ 510^o \\ \\[-3mm] 750^o \\ 870^o\end{Bmatrix}$

$\text{Divide by 3: }\;x \;=\;\begin{Bmatrix}10^o \\ 50^o \\ 130^o \\ 170^o \\ 250^o \\ 290^o \end{Bmatrix}$

And we have all the solution in the interval.

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Am I the only one aware of this?

Am I the only one who teaches this?