# Thread: Solving Trigonometric Equations Exactly

1. ## 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

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

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

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

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

etc.

-Dan

3. Hello, Mantissa!

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

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

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

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: .$\displaystyle \sin60^o \:=\:\frac{\sqrt{3}}{2}$

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

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

4. 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.

5. 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.

6. 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:

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

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

$\displaystyle a^{2} + b^{2} = c^{2}$

7. 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