How to solve 12sinx-1.8cosx=10 without a graphing calculator?

[theBEAST]

I tried squaring both sides but can't seem to simplify it. Any ideas on how this can be solved?

Thanks!

Edit: For those who are interested, I have attached the question that has prompted me to solve this system of equations.

 

[ehild]

Both sinx and cosx can be expressed by tan(x/2)

\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}

\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}


With these substitutions, you get a quadratic equation for tan(x/2).

ehild

 

[vela]

When you have a combination like $A \sin\theta + B \cos\theta$, a useful trick is to write
$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}} \sin\theta + \frac{B}{\sqrt{A^2+B^2}}\cos\theta\right)$$Now identify
\begin{align*}
\cos\phi &= \frac{A}{\sqrt{A^2+B^2}} \\
\sin\phi &= \frac{B}{\sqrt{A^2+B^2}}
\end{align*}then you have
$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}(\cos\phi \sin\theta + \sin\phi\cos\theta) = \sqrt{A^2+B^2}\sin(\theta+\phi)$$where $\tan\phi = B/A$.

In this particular problem, if you move the factor of 4 kN over to the righthand side, you have
$$3 \sin \theta - 0.45 \cos \theta = 2.5$$and $\tan\phi = -0.45/3$, from which you should recognize ɸ as being the angle the line connecting A to the point where the force acts makes with the horizontal.

 

[SammyS]

I tried squaring both sides but can't seem to simplify it. Any ideas on how this can be solved?

Thanks!

Edit: For those who are interested, I have attached the question that has prompted me to solve this system of equations.

What system of equations.

I see none.

 

[ehild]

What system of equations.

I see none.

It is in the title. 12sinx-1.8cosx=10

ehild

 

[SammyS]

It is in the title. 12sinx-1.8cosx=10

ehild

Thanks e .

Any important information in the title should also appear in the body of the message.

Also, OP refers to "... this system of equations".