IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Ask anything and receive thorough, reliable answers from our community of experienced professionals.

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]


Sagot :

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]