Question 250683: could you PLEASE help me solve this trig identity.
(6sin^x)-(5cosx)-(2)=0 and the range is (0,2pi)
im pretty sure i have to factor it but i cant seem to figure out how to do.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
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
an identity, you are saying that the relationship is true  (because that is your given domain). Of course, this statement is utterly false. Consider the counter-example  :
So, what I think you want is to solve the equation:
for all values of  on the interval )
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:
which is just a consequence of Pythagoras applied to the unit circle. This can be written:
And then a suitable substitution can be made in your equation:
\,-\,5\cos{x}\,-\,2\ =\ 0)
A little distribution and simplification yields:
Now, in order that we don't get lost in a bunch of trigonometric functions, let's make a convenient substitution. Let  , and then substitute:
Leaving us with a factorable quadratic.
\left(3u\,+\,4\right)\ =\ 0)
And we then know that  or 
But since , we can say:
 .
However, we cannot say:
because the range of the cosine function is 
Now all that is left is to solve the equation:
)
On the interval
For this we can use a calculator and the inverse cosine function to calculate one of the values. Then subtract that value from  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 
Or perhaps you have simply committed to memory that }\ =\ \cos{\left(\frac{5\pi}{3}\right)}\ =\ \frac{1}{2})
John
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