Find all solutions of the equation in the interval [tex]\([0, 2\pi)\)[/tex].

[tex]\[2 \cos^2 x - 3 \sin x = 3\][/tex]

Write your answer in radians in terms of [tex]\(\pi\)[/tex]. If there is more than one solution, separate them with commas.

[tex]\[x = \boxed{\phantom{}} , \boxed{\phantom{}}\][/tex]



Answer :

To find all solutions of the equation [tex]\( 2 \cos^2 x - 3 \sin x = 3 \)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], follow these steps:

1. Rewrite Using a Trigonometric Identity:
The given equation is [tex]\( 2 \cos^2 x - 3 \sin x = 3 \)[/tex]. We know that [tex]\( \cos^2 x = 1 - \sin^2 x \)[/tex]. Substituting this identity into the equation gives:
[tex]\[ 2 (1 - \sin^2 x) - 3 \sin x = 3 \][/tex]
Simplify this to:
[tex]\[ 2 - 2 \sin^2 x - 3 \sin x = 3 \][/tex]

2. Rearrange the Equation:
Move all terms to one side of the equation:
[tex]\[ -2 \sin^2 x - 3 \sin x + 2 - 3 = 0 \][/tex]
Simplify further:
[tex]\[ -2 \sin^2 x - 3 \sin x - 1 = 0 \][/tex]
Multiply the whole equation by [tex]\(-1\)[/tex] to make it positive:
[tex]\[ 2 \sin^2 x + 3 \sin x + 1 = 0 \][/tex]

3. Solve the Quadratic Equation:
Let [tex]\( u = \sin x \)[/tex]. The quadratic equation now becomes:
[tex]\[ 2u^2 + 3u + 1 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. Here, [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \][/tex]
Simplify under the square root:
[tex]\[ u = \frac{-3 \pm \sqrt{9 - 8}}{4} \][/tex]
[tex]\[ u = \frac{-3 \pm 1}{4} \][/tex]
Thus, we have two solutions for [tex]\( u \)[/tex]:
[tex]\[ u = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \][/tex]
[tex]\[ u = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \][/tex]

4. Find [tex]\( x \)[/tex] using the values of [tex]\( u \)[/tex]:
- For [tex]\( \sin x = -\frac{1}{2} \)[/tex]:
The solutions for [tex]\( x \)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ x = \frac{7\pi}{6}, \frac{11\pi}{6} \][/tex]
- For [tex]\( \sin x = -1 \)[/tex]:
The solution for [tex]\( x \)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] is:
[tex]\[ x = \frac{3\pi}{2} \][/tex]

Combining all solutions, we get:
[tex]\[ x = \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \][/tex]
So, the final answer in radians in terms of [tex]\(\pi\)[/tex] is:
[tex]\[ x = \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \][/tex]