Respuesta :

nareenasad80

Answer:

If you meant 2 sin⁡(2x) = [tex]\frac{1}{sin(x)}[/tex]​, let's solve for x in the given range 0≤x≤[tex]360^{0}[/tex].

Firstly, let's simplify the equation by multiplying both sides by sin(x) to get rid of the fraction:

2 sin⁡(2x) ⋅ sin⁡(x) = 1

Now, use the double-angle identity for sine (sin⁡(2θ) = 2 sin⁡(θ) cos⁡(θ)):

4 sin⁡(x) cos⁡(x) ⋅ sin⁡(x) = 1

Combine terms and rearrange:

4 sin⁡2(x) cos⁡(x) = 1

Now, apply the identity [tex]sin^2[/tex](θ) + [tex]cos^2[/tex](θ) = 1 to replace cos⁡(x):

4(1 − [tex]cos^2[/tex](x)) cos⁡(x) = 1

Expand and rearrange:

4 cos⁡(x) − 4[tex]cos^3[/tex](x) = 1

Move all terms to one side of the equation:

4 [tex]cos^3[/tex](x) + 4cos⁡(x) − 1 = 0

Now, this cubic equation can be solved to find values of cos(x), and then the corresponding values of x. Keep in mind that the solutions within the specified range 0≤x≤[tex]360^{0}[/tex] should be considered.

Otras preguntas