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Math Help - Solving Trigonometric Equations Exactly

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Mantissa View Post
    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
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  3. #3
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    Hello, Mantissa!

    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}}

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  4. #4
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    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.
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  5. #5
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    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.
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  6. #6
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    Quote Originally Posted by Mantissa View Post
    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}
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mantissa View Post
    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
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