September 29, 2020, 01:52:38 am

### AuthorTopic: Solving Trigonometric Equations  (Read 314 times)

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#### K.Smithy

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##### Solving Trigonometric Equations
« on: January 07, 2020, 04:44:27 pm »
+1
Hi

I was just smashing out some maths study and it was going pretty well until I got to some worked examples that have confused me.
My textbook seems to contradict itself with two worked examples, so now I'm unsure of the correct procedure.
The questions are:

Solve the following equations for x
a) 2sin(2x) - 1 = 0, 0 ≤ x ≤ 2π
working out
Change domain
2sin(2x) - 1 = 0, 0 ≤ 2x ≤ 4π

2sin(2x) = 1
sin(2x) = 1/2

(USE SPECIAL TRIANGLE AND UNIT CIRCLE TO FIND EXACT VALUES)
^ the base ended up being π/6

2x = 0 + π/6, π - π/6, 2π + π/6, 3π - π/6
2x = π/6, 5π/6, 13π/6, 17π/6
Get rid of the 2
x = π/12, 5π/12, 13π/12, 17π/12

Note: at the end we got rid of the 2 by applying it to the denominator of the fraction

The next example was:
b) 2cos(2x - π) - 1 = 0, -π ≤ x ≤ π
working out
Change domain
2cos(2x - π) - 1 = 0, -3π ≤ 2x - π ≤ π

2cos(2x - π) = 1
cos(2x - π) = 1/2

(USE SPECIAL TRIANGLE AND UNIT CIRCLE TO FIND EXACT VALUES)
^ the base ended up being π/3

2x - π = -2π - π/3, -2π + π/3, 0 - π/3, 0 + π/3
2x - π = -7π/3, -5π/3, -π/3, π/3
2x = -7π/3 + π, -5π/3 + π, -π/3 + π, π/3 + π
2x = -4π/3, -2π/3, 2π/3, 4π/3
x = -2π/3, -π/3, π/3, 2π/3

Note: at the end we got rid of the 2 by applying it to the numerator of the fraction

So now I am confused as to whether we apply the changes to the denominator or numerator? My textbook changes its mind a couple times and will sometimes apply it to the denominator and other times the numerator.. Is it just a matter of which one is easiest to change? i.e. if you had 2x = 3π/4 you would make it 3π/8 instead of 1.5π/4 because it is "easier"??

Thanks!
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#### RuiAce

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##### Re: Solving Trigonometric Equations
« Reply #1 on: January 07, 2020, 04:51:45 pm »
+4
I mean, the way I read this, I think the source of the problem is your confusion about $\frac{3\pi}{8}$ versus $\frac{1.5\pi}{4}$.

Remember, they are the same number. If you typed $\frac{1.5}{4}$ and $\frac{3}{8}$ into your calculator, you'll get the same number. This is because we can go from $\frac{1.5}{4}$ to $\frac{3}{8}$ by multiplying both the numerator and denominator by 2. Which is equivalent to multiplying the entire expression by 1, and hence does not change the number.

Similarly, we can go from $\frac{3}{8}$ to $\frac{1.5}{4}$ by dividing by 2.

However as a convention, we always like to keep a fraction as a simplified fraction where possible. This is when the fraction is written as $\frac{a}{b}$, where $a$ and $b$ are both integers (whole numbers), but also have no common factors. So for example, rather than $\frac{1.5}{4}$, we prefer to write it as $\frac{3}{8}$ to avoid the decimal. As another example, rather than $\frac{6}{24}$, we prefer to cancel it back down to $\frac14$.

#### K.Smithy

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##### Re: Solving Trigonometric Equations
« Reply #2 on: January 07, 2020, 05:00:06 pm »
+2
I mean, the way I read this, I think the source of the problem is your confusion about $\frac{3\pi}{8}$ versus $\frac{1.5\pi}{4}$.

Remember, they are the same number. If you typed $\frac{1.5}{4}$ and $\frac{3}{8}$ into your calculator, you'll get the same number. This is because we can go from $\frac{1.5}{4}$ to $\frac{3}{8}$ by multiplying both the numerator and denominator by 2. Which is equivalent to multiplying the entire expression by 1, and hence does not change the number.

Similarly, we can go from $\frac{3}{8}$ to $\frac{1.5}{4}$ by dividing by 2.

Wow... holiday brain is getting me real good 😂 Honestly should have seen that, I don't know what to tell you All I can say is I swear I'm not failing methods.

However as a convention, we always like to keep a fraction as a simplified fraction where possible. This is when the fraction is written as $\frac{a}{b}$, where $a$ and $b$ are both integers (whole numbers), but also have no common factors. So for example, rather than $\frac{1.5}{4}$, we prefer to write it as $\frac{3}{8}$ to avoid the decimal. As another example, rather than $\frac{6}{24}$, we prefer to cancel it back down to $\frac14$.

Ok, this clears everything up.

Thank you so much Rui! (lets just pretend I didn't ask this question, hey? )
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#### K.Smithy

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##### Re: Solving Trigonometric Equations
« Reply #3 on: January 07, 2020, 06:31:13 pm »
+1
Hi... me again

In the worked solution pictured above... I can't quite figure out how root2/2 "suggests" π/4 as neither of the special triangles have root 2 and 2 together.

Thanks
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#### RuiAce

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##### Re: Solving Trigonometric Equations
« Reply #4 on: January 07, 2020, 06:35:10 pm »
+3
Hi... me again

(Image removed from quote.)

In the worked solution pictured above... I can't quite figure out how root2/2 "suggests" π/4 as neither of the special triangles have root 2 and 2 together.

Thanks
\begin{align*} \frac{1}{\sqrt2} &= \frac{\sqrt2}{\sqrt2\times \sqrt2}\\ &= \frac{\sqrt2}{2}. \end{align*}
The triangle should give you the value $\frac1{\sqrt2}$. But more often than not, mathematicians seem to like with rational denominators. So sometimes we prefer $\frac{\sqrt2}{2}$ over $\frac1{\sqrt2}$.

Same thing sometimes happens with $\tan \frac\pi6 = \frac1{\sqrt3} = \frac{\sqrt3}{3}$ as well. These are the only ones that require you watch out for 'rationalising the denominator', if I recall correctly.

#### K.Smithy

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##### Re: Solving Trigonometric Equations
« Reply #5 on: January 07, 2020, 06:38:55 pm »
+2
This makes so much sense! You are a legend, thanks again!!
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