SOLUTION: 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.

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): You can put this solution on YOUR website!
You'll have to post the problem correctly.
What do you mean by 6sin^x
-----------
Cheers,
stan H.
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:

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.

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

John

RELATED QUESTIONS
could you PLEASE help me solve this trig identity. (6sinx)-(5cosx)-(2)=0 and the range... (answered by Alan3354)
My question is about functions of standard angles. The question is; Find... (answered by Theo)
Here is an example of a Trig function we are working on that I HAVE been able to solve, I (answered by Alan3354)
Please help me solve this equation: {{{ 4sin^2(x)+7sin(x)=2 }}} with the range of theta... (answered by ikleyn)
Please help me solve this equation: {{{ cos^2(x)+2sin(x)cos(x)-sin^2(x)=0 }}} with the... (answered by Edwin McCravy)
Can you please help me solve this equation: {{{ 4sin^2(x)+7sin(x)=2 }}} with the range of (answered by Fombitz)
Can you help me to solve Trig equations algebraically? 3tan^3(t)-3tan^2(t)-tan(t)+1=0... (answered by Alan3354)
Hi. This is for a summer AP Calculus Assignment. Please help me: {{{ 2cos( x )= sqrt(... (answered by kmadison,solver91311)
If 0 < p < pi/2 and x=6sin p, what is tan p for the indicated trigonometric function: {{{ (answered by DrBeeee)