# Solving Trigonometric Equations Exactly

• June 10th 2008, 01:03 PM
Mantissa
Solving Trigonometric Equations Exactly
Hi, I'm new here. I'm a homeschooled senior in my last leg of pre-graduation math stress. Stupidly, I decided to take a pre-calculus class, and found out exactly how far it is beyond my scope of learning capability the hard way.

Anyway, I need help!

Solve exactly. 4sin^2x -3=0, 0 greater than or equal to x, x less than 2pi.

Sorry for not using the proper signs, I'm not sure where to find them on the keyboard. By "pi", I mean Pi. Not just two random variables strung together.

And please, don't think I haven't done any work on this problem--I've spent the last two days trying to find a page in my textbook that will shed light upon the matter, to no avail. None of the examples quite match the construction of this problem, and after flying through a myriad different starting points, all of which became tangled, I'm still just as lost as if I'd never started. It's the only reason I'm not showing any work. :P
• June 10th 2008, 01:09 PM
topsquark
Quote:

Originally Posted by Mantissa
Hi, I'm new here. I'm a homeschooled senior in my last leg of pre-graduation math stress. Stupidly, I decided to take a pre-calculus class, and found out exactly how far it is beyond my scope of learning capability the hard way.

Anyway, I need help!

Solve exactly. 4sin^2x -3=0, 0 greater than or equal to x, x less than 2pi.

Sorry for not using the proper signs, I'm not sure where to find them on the keyboard. By "pi", I mean Pi. Not just two random variables strung together.

And please, don't think I haven't done any work on this problem--I've spent the last two days trying to find a page in my textbook that will shed light upon the matter, to no avail. None of the examples quite match the construction of this problem, and after flying through a myriad different starting points, all of which became tangled, I'm still just as lost as if I'd never started. It's the only reason I'm not showing any work. :P

$4sin^2(x) - 3 = 0$

For convenience and clarity let $y = sin(x)$. Then the equation becomes
$4y^2 - 3 = 0$

$y = \pm \frac{\sqrt{3}}{2}$

$sin(x) = \pm \frac{\sqrt{3}}{2}$

etc.

-Dan
• June 10th 2008, 01:30 PM
Soroban
Hello, Mantissa!

Quote:

Solve exactly: . $4\sin^2\!x -3\:=\:0,\quad 0 \leq x < 2\pi.$
First, solve for $x$

We have: . $4\sin^2\!x -3 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!x \:=\:3 \quad\Rightarrow\quad \sin^2\!x \:=\:\frac{3}{4}$

Take square roots: . $\sin x \:=\:\pm\sqrt{\frac{3}{4}} \quad\Rightarrow\quad \sin x \:=\:\pm\frac{\sqrt{3}}{2}$

At this point, we are expected to be familiar with some special angles.
. . We should know the trig values for 30°, 60°, and 45°.

And we recognize that: . $\sin60^o \:=\:\frac{\sqrt{3}}{2}$

From there, we know that: . $\sin120^o \:=\:\frac{\sqrt{3}}{2},\;\;\sin240^o\:=\:-\frac{\sqrt{3}}{2},\;\;\sin300^o\:=\:-\frac{\sqrt{3}}{2}$

Therefore: . $x \;\;=\;\;60^o,\:120^o,\:240^o,\:300^o \;\;=\;\;\boxed{\frac{\pi}{3},\:\frac{2\pi}{3},\:\ frac{4\pi}{3},\:\frac{5\pi}{3}\text{ radians}}$

• June 10th 2008, 01:48 PM
Mantissa
Wow. I just had that "OOOOOH" feeling you get when something suddenly clicks.

Thank you both tremendously! Soroban, I totally forgot about radians (which explains why the answer in the book was so weird!) and Dan, that's an excellent way to think of it. So much simpler.

THANK YOU.
• June 11th 2008, 05:35 PM
Mantissa
All right... I need some help again, regarding the same subject.

The problem looks like this:

sin[sin^-1(3/5)+cos^-1(4/5)]

It says to "Solve exactly without the use of a calculator".

What I don't understand, or have forgotten through lack of use, is how you solve a sin or cosine without the use of a calculator? Is this another problem concerning radian measure?

Would it be possible for someone to solve a similar problem to this one, so I can see from the example how it's to be done?

Thank you.
• June 11th 2008, 05:57 PM
TKHunny
Quote:

Originally Posted by Mantissa
All right... I need some help again, regarding the same subject.

The problem looks like this:

sin[sin^-1(3/5)+cos^-1(4/5)]

It says to "Solve exactly without the use of a calculator".

What I don't understand, or have forgotten through lack of use, is how you solve a sin or cosine without the use of a calculator? Is this another problem concerning radian measure?

Would it be possible for someone to solve a similar problem to this one, so I can see from the example how it's to be done?

Thank you.

You should not need a calculator for this one. You need these:

$\sin(a+b)\;=\;\sin(a)\cos(b)\;+\;\cos(a)\sin(b)$

$\sin^{-1}(\sin(a))\;=\;\sin(\sin^{-1}(a))\;=\;a$

$a^{2} + b^{2} = c^{2}$
• June 12th 2008, 03:52 AM
topsquark
Quote:

Originally Posted by Mantissa
All right... I need some help again, regarding the same subject.

The problem looks like this:

sin[sin^-1(3/5)+cos^-1(4/5)]

It says to "Solve exactly without the use of a calculator".

What I don't understand, or have forgotten through lack of use, is how you solve a sin or cosine without the use of a calculator? Is this another problem concerning radian measure?

Would it be possible for someone to solve a similar problem to this one, so I can see from the example how it's to be done?

Thank you.

New questions should go in new threads.

-Dan