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anonymous
 4 years ago
how do i solve: 49.51cosx  30sinx = 1.22625
anonymous
 4 years ago
how do i solve: 49.51cosx  30sinx = 1.22625

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol i'll just let technology do i rather than doing the newtonraphson method. thanks for the help though, i would have never thought of solving it with using the newton raphson method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 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.
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