Trig Identity Help

jesjesjesjes · 2014-03-22 06:31:13

Solve for x: 3cosx + sin2x = 1
0 <= x <= 10

Answer:1.37, 5.44, 7.65

I tried converting sin2x to 2sinx * cosx, but I don't know what to do after that

eigenguy · 2014-03-23 07:00:39

After that square both sides, then make another substitution that will leave you a polynomial equation on just 1 function. Solve the polynomial (easier said than done - unless I made a mistake somewhere it doesn't have any rational roots), then take the inverse trig function to get the answers.

Bob · 2014-03-23 07:54:24

hi eigenguy,

I was just coming to a similar conclusion. I get a quartic with no 'obvious' solutions, so I'm left with trial and improvement or a graphical approach. And if the latter, then why not just graph the original for a solution. I can show there's a single solution in the range 0 to pi/2 at about acos(0.2) without resorting to the above, but that's hardly an 'analytic' approach.

jesjesjesjes: Where did this come from? If it's from a book or a teacher, what was the lesson just before the question was set?

Bob

eigenguy · 2014-03-25 10:20:29

Yeah. When I did it, I accidently dropped the square from the (3+2sin x) factor, and just got a cubic. Still couldn't solve it easily, though.

Given that the answers are in decimal, I wouldn't be surprised if this was intended to be solved numerically. I would suggest using Newton's method rather than trial and error, or even graphical solutions. It might still be easier to use Newton's method to solve the quartic. I doubt Ferrari's method is would be any more illuminating.

Last edited by eigenguy (2014-03-25 10:34:00)

eigenguy · 2014-03-25 16:26:32

The polynomial is f(t) = (3+2t)[sup]2[/sup](1-t[sup]2[/sup]) - 1. It has turning points at roughly -3/2, -7/8, 3/8 (the later two are only approximate). Since t = sin x, it only extends in value from -1 to 1. So the -3/2 turning point is of no interest. f(-1) = f(1) = -1. f(3/8) ~= 12.5, f(-7/8) ~= -6.5. So there are 2 roots of the polynomial in the interval of interest: One between -7/8 and 3/8, and another between 3/8 and 1.

I don't have time now, but you could almost certainly estimate them very closely using Newton's method, then take the inverse sin to find the values of x between 0 and 10 that yield them.

Last edited by eigenguy (2014-03-26 00:29:42)

bobbym · 2014-03-25 19:29:55

From the plot we use

x_i = {1.37, 5.44, 7.65}

also from the plot we can see that the roots are isolated end even Newton's should not have any problems with convergence.

Convergence is rapid and we get

x = 1.36774681885021154561007...

x = 5.43903726392767380439233...

x = 7.65093212602979802253535...

The methods of experimental math can now be tried in an effort to recover the closed forms for the three roots.

Using an Integer Relation Algorithm ( PSLQ ) we recover:

It is highly unlikely that this is what was demanded of the OP. It is more likely that he has copied the problem incorrectly or has made it up. Probably, the problem is the simpler one:

Or perhaps, the teacher or the OP added the 1 to an already solvable problem thinking it would not change the solution very much. This of course is foolish thinking ( see the problem of the counting frog ).

Math Is Fun Forum

#1 2014-03-22 06:31:13

Trig Identity Help

#2 2014-03-23 07:00:39

Re: Trig Identity Help

#3 2014-03-23 07:54:24

Re: Trig Identity Help

#4 2014-03-25 10:20:29

Re: Trig Identity Help

#5 2014-03-25 16:26:32

Re: Trig Identity Help

#6 2014-03-25 19:29:55

Re: Trig Identity Help

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