if x^2+(3*sin(x))-2=0 then what is the value of x?

Let f(x)=x²+3sin(x)-2.  We want the value(s) of x that make f(x)=0.

First let's plot a graph of the function:

We can see there are two solutions to f(x) = 0.  Because of the sin(x) term we need to solve this numerically.  One way of doing this is to use the Newton-Raphson method:

$$\textbf{Newton-Raphson numerical method}.\\ $$ f(x)=x^2+3sin(x)-2 $$\\ $$ f'(x)=2x+3cos(x) $$Guess an initial value for $x$ call it $x_0$. Then iterate towards a solution as follows:$$ x_{n+1}=x_n-\frac{f(x_n}{f'(x_n)} $$\\ Because there are two solutions here, we need to do this twice with a different guess each time. From the graph we can see that $x_0=-2$ and $x_0=1$ might be good guesses.\\ Because Newton-Raphson tends to converge quickly, we probably don't need to iterate many times. Here after about 4 iterations we converge on$$ x=-2.1310748 $$and$$ x=0.5853206 $$from our two guesses, which looks consistent with the graph above!$$

if x^2+(3*sin(x))-2=0 then what is the value of x?

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This one would be difficult to solve without a solving program....although we could use some Calculus techniques to get an approximation, like Alan did. However, since you're asking a trig question, I might assume that you haven't taken Calculus, yet.

Using Wolfram, the solutions (in rads) are -2.13105 (-122.1°) and .585321 (33.54°)

I think that's it.......

Alan, did you do that whole answer in Texmaker and then copy it over?

Alan, did you do that whole answer in Texmaker and then copy it over?

I did (well, the LaTeX bit - not the graph!) and it didn't work perfectly (the first few lines aren't in the same position as in Texmaker)!

Chris wrote; 

This one would be difficult to solve without a solving program....although we could use some Calculus techniques to get an approximation, like Alan did. However, since you're asking a trig question, I might assume that you haven't taken Calculus, yet.

Using Wolfram, the solutions (in rads) are -2.13105 (-122.1°) and .585321 (33.54°)

We should note that Wolfram's results are also approximate.

An alternative, non-calculus based technique, would be a binary section technique.  Choose two values of opposite sign either side of a solution (use the graph to come up with initial choices).  Then halve the interval and see if f(x) is sufficiently close to zero. If it isn't but is positive, replace the previous positive value by this new one (similarly, if the new one is negative, replace the previous negative value by the new one) and keep repeating this process until f(x) is within some pre-specified tolerance of zero.