1. More trig review

Ok I'm having problems with these 3 problems

Solve each equation in the interval {0,2pi) and round your answer to the nearest hundredth of a radian

A) $3\sin^{x} - \sin{x} - 1 =0$

B) $2\sin{x} - 3\cos{x} = 1$

C) $3\sin{x} - 5 = 0$

Ok so for problem A I did this

I used u of sub and solved the quad equation which gave me

$\frac{1 \pm \sqrt{13}}{6} = u$

so

$\sin{x} = \frac{1 \pm \sqrt{13}}{6}$

Now I take arcsin of both sides and get

3.59, .88, 5.83, 2.26

I believe this is correct

For problem B I'm not sure where to start.

I'm guessing either square both sides or equal the right side to cos^2x + sin^2x =1

For problem C

I get that it is false due to range

Any help would be appreciated and thanks in advance.

2. Hi,
Originally Posted by 11rdc11
Ok so for problem A I did this
(...) I believe this is correct
I think it is correct too.

Originally Posted by 11rdc11
C) $3\sin{x} - 5 = 0$
Originally Posted by 11rdc11
For problem B I'm not sure where to start.

I'm guessing either square both sides or equal the right side to cos^2x + sin^2x =1
I'd rather do it this way :

\begin{aligned}
2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\
&=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)
\end{aligned}
This is interesting because $\left(\tfrac{2}{\sqrt{13}}\right)^2+\left(\tfrac{3 }{\sqrt{13}}\right)^2=1$ so there exists $\phi\in[0,2\pi[$ such that $\begin{cases}\sin \phi = \tfrac{3}{\sqrt{13}}\\ \cos \phi=\tfrac{2}{\sqrt{13}} \end{cases}$. This gives us

$
2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)
=\sqrt{13}\sin(x-\phi)$

Can you take it from here ?
Originally Posted by 11rdc11
For problem C

I get that it is false due to range
Yes, this equation has no solution.

3. Originally Posted by flyingsquirrel
Hi,

I think it is correct too.

I'd rather do it this way :

\begin{aligned}
$
2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\
&=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)
\end{aligned}
" alt="
2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\
&=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)
\end{aligned}
" />
This is interesting because $\left(\tfrac{2}{\sqrt{13}}\right)^2+\left(\tfrac{3 }{\sqrt{13}}\right)^2=1$ so there exists $\phi\in[0,2\pi[$ such that $\begin{cases}\sin \phi = \tfrac{3}{\sqrt{13}}\\ \cos \phi=\tfrac{2}{\sqrt{13}} \end{cases}$. This gives us

$
2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)
=\sqrt{13}\sin(x-\phi)
" alt="
2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)
=\sqrt{13}\sin(x-\phi)
" />

Can you take it from here ?

Yes, this equation has no solution.
Thanks but I don't think my lil cousin will understand that method when I try to explain it to him. I approached it like this

square both sides

Then I go ahead and use u of substitation for the cosx and get

and I get 2.43 and 1.26. Then when I check it graphically it tells me my only solution is 1.26 and 3.84. Where am I going wrong and once again thanks in advance.

• and I get 2.43 and 1.26. Then when I check it graphically it tells me my only solution is 1.26 and 3.84. Where am I going wrong and once again thanks in advance.
In fact there are four solutions : 2.43, 1.26, $3.84=-2.43+2\pi$ and $5.02=-1.26+2\pi$ because $\cos x = \alpha \Longleftrightarrow x\equiv\pm\arccos \alpha\pmod{2\pi}$.

You're probably wondering why we have found four solutions whereas graphically you only see two of them. The reason is that we have solved $(2\sin x)^2=(1+3\cos x)^2$ that is to say $2\sin x =\pm(1+3\cos x)$. However we can easily find the solutions of $2\sin x =+(1+3\cos x)$ : these are the solutions of $2\sin x =\pm(1+3\cos x)$ such that $2\sin x$ and $1+3\cos x$ share the same sign...