Trigonometry Equation: 3sin(2θ) - cos(2θ)

Question

How would you solve the following question:

Question Solve this equation for $0°⩽θ⩽180°$ , show your working.

$$3sin(2θ) - cos(2θ) = 0$$

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My solution

Let $x = 2θ$

$3sin(x) - cos(x) = 0$

$3sin(x) = cos(x)$

$3sin(x) = \sqrt{1 - sin^2(x)}$
(since $sin^2(x) + cos^2(x) = 1$, so $cos^2(x) = 1 - sin^2(x)$ and therefore $cos(x) = \sqrt{1 - sin^2(x)}$)

$9sin^2(x) = 1 - sin^2(x)$, $10sin^2(x) = 1$, $sin^2(x) = 1/10$,

$sin(x) = ±\sqrt{\frac{1}{10}}$

So $x = 18.4°$ (to 1 decimal place)

That is, $2θ = 18.4°$ (to 1 decimal place) Therefore $θ = 9.2°$ (to 1 decimal place)

I also thought that another solution was $x = 180° - 18.4° = 161.6°$ (to 1 decimal place) That is, $2θ = 161.6°$ ( to 1 decimal place) Therefore $θ = 80.8°$ (to 1 decimal place)

However, my book is saying that the other solution is $99.2°$ and I don't know how this answer was obtained.

Where have I gone wrong?

Thanks.

Answer 1

$\sin(180^o-x)=\sin(x)$ but $\cos(180^o-x)=\color{red}-\cos(x)$

Answer 2

$\theta = 80.8° $gives$ 2\theta = 161.6°$ so sin is positive but cos is negative, it can't satisfy the given equation. There will be other solution for $sin x= -\sqrt {(1/10)}$ , x= 198.4°. so $\theta = 99.2°$. in this case x= 198.4°, both sin and cos are negative, so it satisfies original equation. (Always try to check the solution by substitution in original equation, in case of trigonometric equations, there can be such deceptive solutions, also check for the domain of tan,cot,sec,cosec!)

Answer 3

It is easier if you divide by $\cos (2\theta)$ to get $$\tan(2\theta)= 1/3$$. Thus you get $\theta = 1/2 \tan^{-1} (1/3)$ and $1/2 \tan^{-1}(1/3)+ \pi/2$

Answer 4

Notice that the sign of $\cos x$ does not matter in $\cos^2 x = 1 - \sin^2 x$. Regardless of the sign of $\cos x$, its square is nonnegative. But then you write $\cos x = \mathbf{+{}}\sqrt{1 - \sin^2 x}$, so you have decided that $\cos x$ can only be positive. What about the other value whose square is $\cos^2 x$, $\cos x = - \sqrt{1 - \sin^2 x}$. Unless you know your angles are in quandrants I and IV, you do not konw that you can ignore the angles with negative cosines.

So from $3 \sin x = \cos x$, you would produce \begin{align*} 3 \sin x &= \sqrt{1 - \sin^2 x} &&\text{ or } & 3 \sin x &= -\sqrt{1 - \sin^2 x} \text{.} \end{align*}

However, generally speaking, introducing radicals that you are going to have to remove by raising to powers is a recipe for producing spurious solutions. (Example of spurious solution. From $x=-1$, we deduce $x^2 = 1$, which has two solutions $x = -1$ (from which we started) and $x = 1$, which appeared when we squared both sides of the equation, but was not a fact from which we started.)

Far better is to use identities and factor.

Answer 5

When $θ=99.2°$, $2θ=198.4°$.

Since $sin(x)$ can be equal to $-\sqrt\frac{1}{10}$, $x$ can be equal to $-18.4°$

And since $sin2θ=sin198.4°=sin(-18.4°)$, you do lack one solution, that is, missing the result from the negative case.