Question 826685
Finding the zeros of
y = 3sin(2x) + cos(x)
means finding the solutions to:
0 = 3sin(2x) + cos(x)<br>
Solving Trig equations like this usually starts with using algebra and/or Trig properties to transform the equation into one or more equations of the general form:
TrigFunction(expression) = number<br>
To solve this equation we will start by using the sin(2x) = 2sin(x)cos(x) identity. Substituting for sin(2x) we get:
0 = 3(2sin(x)cos(x)) + cos(x)
which simplifies to:
0 = 6sin(x)cos(x) + cos(x)
Now we can factor out cos(x):
0 = cos(x)(6sin(x)+1)
From the Zero Product Property:
cos(x) = 0 or 6sin(x)+1 = 0
(Note how we now have two equations. This is one way to get more than one equation from a single equation.) Solving these we get:
cos(x) = 0 or sin(x) = -1/6
We now have the equations in the desired form.<br>
Next we find the general solution.
For cos(x) = 0 we should recognize that 0 is a special angle value for cos. We should know that cos is 0 at {{{pi/2}}}, {{{3pi/2}}} and any other angle which is co-terminal with these. So the general solution for cos(x) = 0 is:
{{{x = pi/2 + 2pi*n}}}
{{{x = 3pi/2 + 2pi*n}}}<br>
For sin(x) = -1/6 we should recognize that 1/6 is not a special angle value for sin. So we will need our calculators. We will use {{{sin^(-1)(1/6)}}} to find the reference angle. (Note: Do not use negative values when finding reference angles!) With my calculator set to radian mode, I get a reference angle, rounded to 4 places, of 0.1674. Since the 1/6 was negative (<i>Here</i> is where we use the fact that the sin is negative!) and since sin is negative in the 3rd and 4th quadrants we get the following general solution equations for sin(x) = -1/6:
{{{x = pi + 0.1674 + 2pi*n}}} for the 3rd quadrant
{{{x = -0.1674 + 2pi*n}}} for the 4th quadrant<br>
Together, the general solution equations for 0 = 3sin(2x) + cos(x) are:
{{{x = pi/2 + 2pi*n}}}
{{{x = 3pi/2 + 2pi*n}}}
{{{x = pi + 0.1674 + 2pi*n}}}
{{{x = -0.1674 + 2pi*n}}}<br>
The general solution equations express the infinite set of solutions to the equation. Many problems, like this one, ask for solutions which are in a given interval. This time the interval is {{{-pi <= x <= pi}}} To find these answers we replace the n's in the equations with various integers until we have found all the x's which are in the interval.<br>
From {{{x = pi/2 + 2pi*n}}}...
If n = 0 then {{{x = pi/2}}}
If n = 1 (or greater) then x is too large for the interval.
If n = -1 (or smaller) then x is too small for the interval.<br>
From {{{x = 3pi/2 + 2pi*n}}}...
If n = 0 (or larger) then x is too large for the interval.
If n = -1 then {{{x = -pi/2}}}
If n = -2 or smaller then x is too small for the interval.<br>
From {{{x = pi + 0.1674 + 2pi*n}}}...
If n = 0 (or larger) then x is too large for the interval.
If n = -1 then {{{x = -pi+0.16674}}}
If n = -2 or smaller then x is too small for the interval.<br>
From {{{x = -0.1674 + 2pi*n}}}...
If n = 0 then {{{x = -0.1674}}}
If n = 1 (or greater) then x is too large for the interval.
If n = -1 (or smaller) then x is too small for the interval.<br>
Altogether, the solutions to 0 = 3sin(2x) + cos(x) (and therefore the zeros of f(x) = 3sin(2x) + cos(x)) within the specified interval are:
{{{x = pi/2}}}, {{{x = -pi/2}}}, {{{x = -pi+0.1674}}} and {{{x = -0.1674}}}<br>
P.S. Feel free to replace the {{{pi}}} in {{{-pi+0.1674}}} with a decimal approximation and then add the two decimals.