Question 815618
NOTE: My original posting had an error. This posting has corrected this.<br>
{{{12sin^2(w)+13cos(w)-15=0}}}
If we could factor this we might be able to solve it. But with both sin and cos in the equation it will not factor easily. Since the sin is squared, we can use {{{sin^2(x)=1-cos^2(x)}}} to substitute in for the sin squared:
{{{12(1-cos^2(w))+13cos(w)-15=0}}}
which simplifies as follows:
{{{12-12cos^2(w)+13cos(w)-15=0}}}
{{{-12cos^2(w)+13cos(w)-3=0}}}
Now we can factor. Since factoring with a positive leading coefficient is easier, I'll start by multiplying (or dividing) both sides by -1:
{{{12cos^2(w)-13cos(w)+3=0}}}
Factoring we get:
(3cos(w)-1)(4cos(w)-3) = 0
From the Zero Product Property:
3cos(w) - 1 = 0 or 4cos(w)-3 = 0
Solving 3cos(w) - 1 = 0:
3cos(w) = 1
cos(w) = 1/3
Using inverse cos we get a reference angle of (approximately) 1.23. Since the 1/3 is positive and since cos is positive in the 1st and 4th quadrants we get general solution equations of:
{{{w = 1.23 + 2pi*n}}} (for 1st quadrant angles)
{{{w = -1.23 + 2pi*n}}} (for 4th quadrant angles)<br>
Solving 4cos(w)-3 = 0:
4cos(w) = 3
cos(w) = 3/4
Using inverse cos we get a reference angle of (approximately) 0.72. Since the 1/3 is positive and since cos is positive in the 1st and 4th quadrants we get general solution equations of:
{{{w = 0.72 + 2pi*n}}} (for 1st quadrant angles)
{{{w = -0.72 + 2pi*n}}} (for 4th quadrant angles)<br>
Now we use the general solution equations to find the specific solutions which are in the given interval.
From {{{w = 1.23 + 2pi*n}}} we get:
if n = 0 then w = 1.23
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From {{{w = -1.23 + 2pi*n}}} we get:
if n = 0 or smaller then w is too small for the interval
if n = 1 then w = -1.23 + 6.28 = 5.05
if n = 2 (or larger) then w is too large for the interval
From {{{w = 0.72 + 2pi*n}}} we get:
if n = 0 then w = 0.72
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From {{{w = -0.72 + 2pi*n}}} we get:
if n = 0 or smaller then w is too small for the interval
if n = 1 then w = -0.72 + 6.28 = 5.56
if n = 2 (or larger) then w is too large for the interval<br>
So the only solutions in the given interval are: 1.23, 5.05, 0.72, 5.56