Csdrhrt

Solve the trigonometric equation $sin 2x+cos x=0$

My Attempt
$$
2sin xcos x+cos x=0impliescos x=0 text{ or }sin x=frac{-1}{2}=sinfrac{-pi}{6}\
x=(2n+1)frac{pi}{2} text{ or }x=npi+(-1)^nfrac{-pi}{6}\
x=npi+frac{pi}{2} text{ or }x=2npi-frac{pi}{6}text{ or }x=2npi+pi+frac{pi}{6}
$$
Reference
$$
cos x=-sin 2x=cosBig(frac{pi}{2}+2xBig)implies x=2npipmBig(frac{pi}{2}+2xBig)\
-x=2npi+frac{pi}{2}text{ or }3x=2npi-frac{pi}{2}implies x=2mpi-frac{pi}{2}text{ or }x=frac{2mpi}{3}-frac{pi}{6}
$$
But my reference gives the solution $x=2npi-dfrac{pi}{2}$ or $x=dfrac{2npi}{3}-dfrac{pi}{6}$. I understand how it is reached and both represent the same solutions. But, how do I derive the solution in my reference from what I found in my attempt ? i.e.,

How to derive
$$
bigg[x=npi+frac{pi}{2} text{ or }x=2npi-frac{pi}{6}text{ or }x=2npi+pi+frac{pi}{6}bigg]\
implies bigg[x=2npi-dfrac{pi}{2}text{ or }x=dfrac{2npi}{3}-dfrac{pi}{6}bigg]
$$

There are two ways to solve the equation.

Your method: $2sin xcos x+cos x=0$, so $cos x(2sin x+1)=0$. Thus we have either $cos x=0$ or $sin x=-1/2$. Thus
begin{align}
x&=frac{pi}{2}+2npi &text{or}&& x&=-frac{pi}{2}+2npi && text{(from $cos x=0$)} \[4px]
x&=-frac{pi}{6}+2npi &text{or}&& x&=frac{7pi}{6}+2npi && text{(from $sin x=-1/2$)}
end{align}
(you grouped together the families of solutions of $cos x$ and of $sin x=-1/2$, but I prefer to keep them separated). Good job.

Alternative method: $cos x=-sin2x=cos(frac{pi}{2}+2x)$. Therefore either
$$
x=frac{pi}{2}+2x+2npi to x=-frac{pi}{2}-2npi
$$
or
$$
x=-frac{pi}{2}-2x+2npi to 3x=-frac{pi}{2}+2npi to x=frac{2npi}{3}-frac{pi}{6}
$$

How do you recover the previous sets of solutions?

One set is already present. For the other three, consider the cases when $n=3k$, $n=3k+1$ or $n=3k+2$, with integer $k$. Then
begin{align}
n&=3k & x&=frac{6kpi}{3}-frac{pi}{6}=2kpi-frac{pi}{6} \[4px]
n&=3k+1 & x&=frac{(6k+2)pi}{3}-frac{pi}{6}=2kpi+frac{2pi}{3}-frac{pi}{6}=2kpi+frac{pi}{2}\[4px]
n&=3k+2 & x&=frac{(6k+4)pi}{3}-frac{pi}{6}=2kpi+frac{4pi}{3}-frac{pi}{6}=2kpi+frac{7pi}{6}
end{align}

You and your reference (assuming no typos) have omitted many solutions. For instance, $x = pi/2$, which you are missing, and $-pi/2$, which your reference is missing.

Assuming throughout that $k$ is any integer.

From $cos(x) = 0$, $x = pm cos^{-1}(0) + 2pi k$, giving the two solution families $x = pm frac{pi}{2} + 2pi k$. These can be written as a single family (note the coefficient of $k$): $frac{pi}{2} + pi k$, which is equivalent to $frac{-pi}{2} + pi k$. This equivalence is most likely what is going on between your solution and your reference's.

From $sin(x) = frac{-1}{2}$, $x = sin^{-1}left(frac{-1}{2}right) + 2pi k = frac{-pi}{6} + 2 pi k$ or $x = pi - sin^{-1}left(frac{-1}{2}right) + 2pi k = frac{7 pi}{6} + 2 pi k$. Taken together, you should recognize these as your solution and one of them you quote from your reference.

Are you sure you have copied your reference's entire answer and correctly copied its $k$ coefficients?

We have that

• $cos x = 0 implies x=frac{pi}2+kpi$




• $sin x = -frac12 implies x=-frac{pi}6+2kpi quad lor quad x=-frac{5pi}6+2kpi$

which is equivalent to

• $x=-frac{pi}2+2kpi$


• $x=-frac{pi}6+frac23kpi$

to see that draw the solution points on the trigonometric circle.

$$sin(2x)+cos(x)=0$$
$$2sin(x)cos(x)+cos(x)=0$$
$$cos(x)left(2sin(x)+1right)=0$$
so you have two sets of solutions:
$$cos(x)=0,,sin(x)=-frac{1}{2}$$
EDIT:

firstly they have:
$$x=2npi-pi/2=pi(2n-1/2)$$
and you have:
$$x=(2n+1)pi/2$$
let:
$$pi/2(4n-1)=pi/2(2m+1)$$
so:
$$4n-1=2m+1$$
$$m=(4n-2)/2=2n-1$$
so for all integer values of $n$, $m$ is also an integer and so they are equivalent?