Ask your own question, for FREE!
Engineering 13 Online
OpenStudy (anonymous):

how do i solve: 49.51cosx - 30sinx = 1.22625

OpenStudy (anonymous):

There are a number of ways, some better than others. The easiest is probably graphing it using a graphing calculator and looking for the points at which is crosses the x axis, these will be your possible values for x. You could also use a solve function on your calculator (assuming it has one). Plugging the equation into Wolfram Alpha would also work. Or you could use something like Newton's Method to get a very close approximate solution. Newton's method will be the most difficult of what I've mentioned, but learning how to do it will also be the most useful in the long run. Of course you do need to know how to take the derivative of the function in order to use Newton's method. If you haven't taken calculus yet, it won't help you.

OpenStudy (anonymous):

lol i'll just let technology do i rather than doing the newton-raphson method. thanks for the help though, i would have never thought of solving it with using the newton raphson method

OpenStudy (anonymous):

the general procedure for solving a*sinx+b*cosx=c is,\[\frac a{\sqrt{a^2+b^2}}sinx+\frac b{\sqrt{a^2+b^2}}cosx=\frac c{\sqrt{a^2+b^2}}.......(1) \]now let,\[\frac a{\sqrt{a^2+b^2}}=\sin{\alpha}, \therefore \frac b{\sqrt{a^2+b^2}}=\cos{\alpha}\] now from (1) we get,\[\sin{\alpha}sinx+\cos{\alpha}cosx=\frac c{\sqrt{a^2+b^2}}\]\[\cos(x-\alpha)=\frac c{\sqrt{a^2+b^2}}\]so, \[x=\alpha +\cos^{-1}(\frac c{\sqrt{a^2+b^2}})\]provided \[|c|\leq \sqrt{a^2+b^2}\]otherwise no solution possible.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!