Question

3arccos (2x+3)=5pi/2

Answer 1

$3arccos(2x+3)=\frac{5\pi }{2}\; \; \to \; \; arccos(2x+3)=\frac{5\pi }{6}\\\\cos(arccos(2x+3))=cos\frac{5\pi }{6}\\\\2x+3=cos(\pi -\frac{\pi}{6})\\\\2x+3=-cos\frac{\pi }{6}\\\\2x+3=-\frac{\sqrt3}{2}\\\\2x=-3-\frac{\sqrt3}{2}=-\frac{6+\sqrt3}{2}\\\\x=-\frac{6+\sqrt3}{4}$

Answer 2

To solve the equation 3arccos(2x + 3) = 5π/2, we can start by isolating the arccos term and then find the value of x.

Let's go step by step:

1. Divide both sides of the equation by 3: arccos(2x + 3) = (5π/2) / 3

2. Simplify the right side: arccos(2x + 3) = (5π/6)

3. Apply the inverse cosine function (arccos) to both sides to eliminate the arccos term: 2x + 3 = cos((5π/6))

4. Evaluate the cosine of (5π/6): cos(5π/6) = -√3/2

5. Rewrite the equation: 2x + 3 = -√3/2

6. Subtract 3 from both sides: 2x = -√3/2 - 3

7. Simplify the right side: 2x = (-√3 - 6)/2

8. Divide both sides by 2: x = (-√3 - 6)/4

So, the solution to the equation 3arccos(2x + 3) = 5π/2 is: x = (-√3 - 6)/4