Cos2x-3cosx+1=cosecx/(cotx-2cot2x) please find general solution of this trigonometric equation

To solve the trigonometric equation \( \cos(2x) - 3\cos(x) + 1 = \frac{\csc(x)}{\cot(x) - 2\cot^2(x)} \), we can break it down step by step. First, we need to simplify both sides of the equation and find a way to express everything in terms of sine and cosine. This will allow us to find the general solution more easily.

Step 1: Simplify the Right Side

We start with the right-hand side of the equation. Recall that:

• \(\csc(x) = \frac{1}{\sin(x)}\)
• \(\cot(x) = \frac{\cos(x)}{\sin(x)}\)

Thus, we can rewrite \(\cot(x) - 2\cot^2(x)\) as follows:

• \(\cot(x) - 2\cot^2(x) = \frac{\cos(x)}{\sin(x)} - 2\left(\frac{\cos(x)}{\sin(x)}\right)^2\)

This simplifies to:

• \(\frac{\cos(x)}{\sin(x)} - 2\frac{\cos^2(x)}{\sin^2(x)}\)
• \(= \frac{\cos(x)\sin(x) - 2\cos^2(x)}{\sin^2(x)}\)

Now, substituting this back into the equation gives us:

\(\cos(2x) - 3\cos(x) + 1 = \frac{1}{\sin(x)} \cdot \frac{\cos(x)\sin(x) - 2\cos^2(x)}{\sin^2(x)}\)

Step 2: Use Trigonometric Identities

Next, we can apply the double angle identity for cosine:

• \(\cos(2x) = 2\cos^2(x) - 1\)

Substituting this identity into the left-hand side of the equation, we have:

\(2\cos^2(x) - 1 - 3\cos(x) + 1 = 2\cos^2(x) - 3\cos(x)\)

Step 3: Set Up the Equation

Now we need to equate both sides:

\(2\cos^2(x) - 3\cos(x) = \frac{\cos(x)\sin(x) - 2\cos^2(x)}{\sin^2(x)}\)

Multiplying both sides by \(\sin^2(x)\) to eliminate the fraction gives us:

\(2\cos^2(x)\sin^2(x) - 3\cos(x)\sin^2(x) = \cos(x)\sin(x) - 2\cos^2(x)\)

Step 4: Rearranging the Equation

Bringing all terms to one side leads us to:

\(2\cos^2(x)\sin^2(x) - 3\cos(x)\sin^2(x) - \cos(x)\sin(x) + 2\cos^2(x) = 0\)

This is a quadratic equation in terms of \(\cos(x)\). We can let \(y = \cos(x)\). Thus:

\(2y^2\sin^2(x) - 3y\sin^2(x) - y\sin(x) + 2y^2 = 0\)

Step 5: Solve the Quadratic Equation

Now we can arrange and solve for \(y\) using the quadratic formula. The standard form is \(Ay^2 + By + C = 0\):

• \(A = 2\sin^2(x)\)
• \(B = -3\sin^2(x) - \sin(x)\)
• \(C = 2\sin^2(x)\)

Using the quadratic formula \(y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\), we can find the values of \(y\) (which is \(\cos(x)\)).

Step 6: Finding General Solutions

Once we have the values for \(y\), we can find \(x\) by using the inverse cosine function. Remember that cosine is periodic, so the general solution will be of the form:

\(x = \pm \cos^{-1}(y) + 2n\pi\) for any integer \(n\).

Final Thoughts

While this gives us a structured approach to finding the general solution, the computation of the quadratic roots and further simplification may require a deeper dive into specific values or numerical methods. The interplay of identities and algebraic manipulation is key here, and practicing with various trigonometric equations will strengthen your skills further.