The number of solutions of the equation sin2x−2cosx+4sinx=4 in the interval [0,5π] is:

To solve this problem, we need to simplify the equation and find the solutions in the given interval [0,5π].

Step 1: Rewrite the equation in terms of sine and cosine The given equation is sin2x - 2cosx + 4sinx = 4. We can rewrite sin2x as 1 - cos^2x using the Pythagorean identity. So, the equation becomes 1 - cos^2x - 2cosx + 4sinx = 4.

Step 2: Simplify the equation Subtract 1 from both sides to get -cos^2x - 2cosx + 4sinx = 3.

Step 3: Express the equation in terms of one trigonometric function We can express cosx in terms of sinx using the identity cosx = sqrt(1 - sin^2x). So, the equation becomes - (sqrt(1 - sin^2x))^2 - 2sqrt(1 - sin^2x) + 4sinx = 3. Simplify this to get -1 + sin^2x - 2sqrt(1 - sin^2x) + 4sinx = 3.

Step 4: Solve the equation This is a quadratic equation in terms of sqrt(1 - sin^2x) and sinx. You can solve it using the quadratic formula or by factoring.

Step 5: Check the solutions in the given interval After solving the equation, you need to check if the solutions are in the interval [0,5π].

Note: This is a complex trigonometric equation and might not have solutions that can be easily found without the use of a calculator or computer software. The exact solutions will depend on the specific equation after simplification.