Question 250683
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In the first place, what makes you think this is an identity?  A trigonometric identity (or any identity, for that matter) is a relationship that is true for all elements of the domain of the expression.


So, by calling


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\sin^2{x}\,-\,5\cos{x}\,-\,2\ =\ 0]


an identity, you are saying that the relationship is true *[tex \LARGE \forall\,x\,|\,x\,\in\,R,\ 0\,<\,x\,<\,2\pi] (because that is your given domain).  Of course, this statement is utterly false.  Consider the counter-example *[tex \LARGE x\ =\ \frac{\pi}{4}]:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,-\,\frac{5\sqrt{2}}{2}\,-\,2\ \neq\ 0]


So, what I think you want is to solve the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\sin^2{x}\,-\,5\cos{x}\,-\,2\ =\ 0]


for all values of *[tex \LARGE x] on the interval *[tex \LARGE \left(0\,<\,x\,<\,2\pi\right)]


The first hurdle to overcome is the fact that the first term involves the sine function and the second term involves the cosine function.  However, there is an identity (and this truly is an identity) that relates the square of the sine and the square of the cosine, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2{x}\ +\ \sin^2{x}\ =\ 1]


which is just a consequence of Pythagoras applied to the unit circle.  This can be written:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2{x}\ =\ 1\ -\ \cos^2{x}]


And then a suitable substitution can be made in your equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\left(1\ -\ \cos^2{x}\right)\,-\,5\cos{x}\,-\,2\ =\ 0]


A little distribution and simplification yields:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\cos^2{x}\,+\,5\cos{x}\,-\,4\ =\ 0]


Now, in order that we don't get lost in a bunch of trigonometric functions, let's make a convenient substitution. Let *[tex \Large u\ =\ \cos{x}], and then substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6u^2\,+\,5u\,-\,4\ =\ 0]


Leaving us with a factorable quadratic.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2u\,-\,1\right)\left(3u\,+\,4\right)\ =\ 0]


And we then know that *[tex \Large u\ =\ \frac{1}{2}] or *[tex \Large u\ =\ \frac{-4}{3}]


But since *[tex \Large u\ =\ \cos{x}], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{x}\ =\ \frac{1}{2}].


However, we cannot say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{x}\ =\ \frac{-4}{3}]


because the range of the cosine function is *[tex \LARGE -1\,\leq\,\cos{x}\,\leq\,1]


Now all that is left is to solve the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \cos^{-1}\left(\frac{1}{2}\right)]


On the interval *[tex \LARGE \left(0\,<\,x\,<\,2\pi\right)]


For this we can use a calculator and the inverse cosine function to calculate one of the values.  Then subtract that value from *[tex \LARGE 2\pi] to get the other one.


Or you can just look at the following unit circle representation to find both of the angles which have intersection points with the circle such that *[tex \LARGE x\ =\ \frac{1}{2}]


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


Or perhaps you have simply committed to memory that *[tex \LARGE \cos{\left(\frac{\pi}{3}\right)}\ =\ \cos{\left(\frac{5\pi}{3}\right)}\ =\ \frac{1}{2}]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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