# Math Help - Need a quick answer check - Trig.

1. ## Need a quick answer check - Trig.

Hello, I am doing trigonometry, and would appreciate a quick answer check. I'll explain what is throwing me off after.

Solve 6sin^2x+sinx-1=0 to nearest hundredth of a radian for (0, 2pi).

Let k=sinx

6k^2+k-1=0
(2k+1)(3k-1) or
(2sinx+1)(3sinx-1)

2sinx+1=0
sinx=(-1/2)
Focus angle is sinx=(1/2) = (pi/6)

Since sinx is negative, find the angle in quadrants three and four.

pi + (pi/6) = 7(pi)/6 and 2pi - (pi/6) = 11(pi)/6

So,

7(pi)/6
11(pi)/6

3sinx-1=0
sinx=(1/3)
Focus angle is sinx=(1/3)

Since sinx is positive, find the angle in quadrants one and two.

x=sin^-1(1/3)
x=19.47 or 19.47(pi)/180 = 0.34 r

(pi)-(19.47(pi)/180) = 160.53(pi)/180 = 2.80 r

11(pi)/6 OR 5.76r
7(pi)/6 OR 3.67r
19.47(pi)/180 OR 0.34r
160.53(pi)/180 OR 2.80r

Now the thing that confuses me is, we know that this is a wave, and we are finding all points on the wave for x when y=0 (x-intercepts). If this is true, would there not be a common difference between two of the values for x? Or does this only apply when the wave has the point of origin as one of its x-values?

Thanks for the help again, and sorry I'm not using proper notation, I'm new to these boards.

1) The period of the sine function is 2pi. Results will ALWAYS be grouped in some way so that infintely many solutions will be exactly 2pi apart. There may be more than one group.

2) There are other symmetric relationships that must be kept in mind. One such is sin(x) = sin(pi - x). Results related in this way will NOT differ by 2pi.

3. Thanks for the tip!

I am now doing some trig identities and would appreciate a quick browse to see if I have made any errors.

I will try to use proper notation to make it easier.

Simplify each trigonometric expression.

a) $\frac{sinx}{tanxcosx}$

$\frac{sinxcosx}{sinxcosx}$

$=1$

b) $sin^4a+sin^2acos^2a$

$=sin^4a+sin^2a(1-sin^2a)$

$=sin^4a+sin^2a-sin^4a$

$=sin^4a-sin^4a+sin^2a$

$=sin^2a$

c) $\frac{sinx-sinxcos^2x}{sin^2x}$

$\frac{sinx-sinx(1-sin^2x)}{sin^2x}$

$\frac{sinx-sinx+sin^2x)}{sin^2x}$

$\frac{sin^2x}{sin^2x}$

$=1$

Sorry if it's still messy, I don't know any of the special commands. I appreciate any help!

EDIT I tried to fix the notation, there is a step missing in part a) but I couldn't figure out how to do a fraction in the denominator.

4. Sorry to make a reply in my own thread but I have tried to fix the notation, for those of you that looked at it before, you probably couldn't even read it well.

Hopefully this makes it easier, I just want a quick yay or nay in the solutions are correct. Thank you!

5. Originally Posted by Jeev
Thanks for the tip!

I am now doing some trig identities and would appreciate a quick browse to see if I have made any errors.

I will try to use proper notation to make it easier.

Simplify each trigonometric expression.

a) $\frac{sinx}{tanxcosx}$

$\frac{sinxcosx}{sinxcosx}$

$=1$

b) $sin^4a+sin^2acos^2a$

$=sin^4a+sin^2a(1-sin^2a)$

$=sin^4a+sin^2a-sin^4a$

$=sin^4a-sin^4a+sin^2a$

$=sin^2a$

c) $\frac{sinx-sinxcos^2x}{sin^2x}$

$\frac{sinx-sinx(1-sin^2x)}{sin^2x}$

$\frac{sinx-sinx+sin^2x)}{sin^2x}$

$\frac{sin^2x}{sin^2x}$

$=1$

Sorry if it's still messy, I don't know any of the special commands. I appreciate any help!

EDIT I tried to fix the notation, there is a step missing in part a) but I couldn't figure out how to do a fraction in the denominator.
If someone could verify my answers

6. I just have to rib you a little. For a), is that expression ALWAYS one (1)? Always?! What about $x = 1$ or $x = \frac{\pi}{2}$?

Just to see if it is soaking in, how about c). Is that ALWAYS one (1)? What about $x = 0$?

Can you prove it one way or the other?

Note: Good work on the LaTeX. Learning to communicate better ALWAYS is worth it, even if x = 0.