Question

6cos^2x+sin^2x-5sinxcosx=0

Answer 1

)))))))))))))))))))))))))))))))))))) @@@@@@@@

Answer 2

We can simplify the given equation by using the identity:

cos^2x + sin^2x = 1

First, we'll move the term with sinxcosx to the right-hand side:

6cos^2x + sin^2x = 5sinxcosx

Now, we'll substitute cos^2x = 1 - sin^2x:

6(1 - sin^2x) + sin^2x = 5sinxcosx

Simplifying this expression, we get:

6 - 6sin^2x + sin^2x = 5sinxcosx

Combining like terms:

6 - 5sin^2x = 5sinxcosx

Now, we can use the identity sin2x = 2sinx cosx:

6 - 5sin^2x = 2sinx(6cosx)

Dividing both sides by 2sinx:

3/sinx - 5sinx = 6cosx

Using the identity 1/sinx = cscx:

cscx - 5sinx = 6cosx

Squaring both sides:

csc^2x - 10sinxcscx + 25sin^2x = 36cos^2x

Substituting cos^2x = 1 - sin^2x:

csc^2x - 10sinxcscx + 25sin^2x = 36 - 36sin^2x

Rearranging terms:

36sin^2x - 10sinxcscx - 36 + csc^2x = 0

Multiplying both sides by sin^2x:

36sin^4x - 10sin^3xcscx - 36sin^2x + sin^2x csc^2x = 0

Substituting 1/sin^2x for csc^2x:

36sin^4x - 10cosx - 36sin^2x + cosx/sin^2x = 0

Multiplying both sides by sin^2x:

36sin^6x - 10cosxsin^2x - 36sin^4x + cosx = 0

Using the identity cosx = sqrt(1 - sin^2x):

36sin^6x - 10sqrt(1 - sin^2x)sin^2x - 36sin^4x + sqrt(1 - sin^2x) = 0

This equation can be difficult to solve algebraically, but we can use numerical methods to find an approximate solution. For example, we could use the Newton-Raphson method to find a root.