# Need some help with solving equations

• Nov 18th 2007, 10:55 AM
finalfantasy
Need some help with solving equations
I am having a lot of trouble with these questions.

6. 5(sin 6t - 4cos 4t) = 0

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10. a) 3sin^2x - 8sinx - 3 = 0

This one I kinda get, but I am having some trouble with it.

The factors can be found like this right?

so therefore ... 3sinx + 1 =0 and sinx - 3 =0

For 3sinx+1=0, you move 1 over, so it becomes 3sinx = -1

You then divide by 3 so it equals sinx = -1/3

it then becomes x = sin^-1(-1/3) but that's as far as I know how to do it.

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I have the answers to these, I just need to know how you get each step to solve the answer.

Also, I have a question concerning angles. For example, would -20 deg, be the same as 20 deg and how so?
• Nov 18th 2007, 12:18 PM
Soroban
Hello, finalfantasy!

Quote:

$\displaystyle 10\;a)\;\;3\sin^2\!x - 8\sin x - 3 \:=\: 0$

The factors can be found like this, right? .$\displaystyle (3\sin x + 1)(\sin x -3) \:=\:0$ . . . . Right!

So therefore: .[1] $\displaystyle 3\sin x + 1 \:=\:0$ .and .[2] $\displaystyle \sin x - 3 \:=\:0$ . . . . Yes!

For [1]: .$\displaystyle 3\sin x + 1\:=\:0$, you move 1 over,
. . so it becomes: .$\displaystyle 3\sin x \: = \:-1$
. . You then divide by 3, so it equals: .$\displaystyle \sin x \:= \:-\frac{1}{3}$
. . It then becomes: .$\displaystyle x \:=\:\sin^{-1}\left(-\frac{1}{3}\right)$ . . . . Good!

but that's as far as I know how to do it.

Now do the same for [2]: .$\displaystyle \sin x - 3 \:=\:0$

Move the -3 over: .$\displaystyle \sin x \:=\:3$

And this equation has no solution.
. . [$\displaystyle \sin x$ is always between -1 and +1.]

Therefore, your answer is the only solution: .$\displaystyle x \;=\;\sin^{-1}\left(-\frac{1}{3}\right)$