Solving Trigonometric Equations

[K.Smithy]

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!

[RuiAce]

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]

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? )

[K.Smithy]

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

[RuiAce]

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]

This makes so much sense! You are a legend, thanks again!!