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Sagot :
Certainly! Let's solve the equation [tex]\(4 \sqrt{3} \csc x - 4 - 3 \csc^2 x = 0\)[/tex] for [tex]\(x\)[/tex] in the interval [tex]\([0, \frac{\pi}{2}]\)[/tex].
### Step 1: Introducing a substitution
Let's introduce a substitution to simplify our work. Let [tex]\(y = \csc x\)[/tex]. Given this substitution, recall that [tex]\(\csc x = \frac{1}{\sin x}\)[/tex]. Thus, the original equation becomes:
[tex]\[ 4 \sqrt{3} y - 4 - 3 y^2 = 0 \][/tex]
### Step 2: Rearrange the equation
Rearrange the equation into standard quadratic form [tex]\(ay^2 + by + c = 0\)[/tex]:
[tex]\[ -3 y^2 + 4 \sqrt{3} y - 4 = 0 \][/tex]
Multiply through by [tex]\(-1\)[/tex] to simplify:
[tex]\[ 3 y^2 - 4 \sqrt{3} y + 4 = 0 \][/tex]
### Step 3: Use the quadratic formula
The quadratic formula states that for an equation [tex]\(ay^2 + by + c = 0\)[/tex], the solutions for [tex]\(y\)[/tex] are given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 3\)[/tex], [tex]\(b = -4\sqrt{3}\)[/tex], and [tex]\(c = 4\)[/tex]. Plugging these into the quadratic formula, we get:
[tex]\[ y = \frac{-(-4\sqrt{3}) \pm \sqrt{(-4\sqrt{3})^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{4\sqrt{3} \pm \sqrt{48 - 48}}{6} \][/tex]
[tex]\[ y = \frac{4\sqrt{3} \pm \sqrt{0}}{6} \][/tex]
[tex]\[ y = \frac{4\sqrt{3}}{6} \][/tex]
[tex]\[ y = \frac{2\sqrt{3}}{3} \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Recall that [tex]\(y = \csc x = \frac{1}{\sin x}\)[/tex]. So we have:
[tex]\[ \csc x = \frac{2\sqrt{3}}{3} \][/tex]
This implies:
[tex]\[ \sin x = \frac{3}{2\sqrt{3}} \][/tex]
[tex]\[ \sin x = \frac{\sqrt{3}}{2} \][/tex]
### Step 5: Determine [tex]\(x\)[/tex] in the given interval
We need to find [tex]\(x\)[/tex] such that [tex]\(\sin x = \frac{\sqrt{3}}{2}\)[/tex] within the interval [tex]\([0, \frac{\pi}{2}]\)[/tex]. We know from trigonometry that:
[tex]\[ x = \frac{\pi}{3} \][/tex]
### Conclusion
Thus, the solution to the equation [tex]\(4 \sqrt{3} \csc x - 4 - 3 \csc^2 x = 0\)[/tex] within the interval [tex]\([0, \frac{\pi}{2}]\)[/tex] is:
[tex]\[ x = \frac{\pi}{3} \][/tex]
### Step 1: Introducing a substitution
Let's introduce a substitution to simplify our work. Let [tex]\(y = \csc x\)[/tex]. Given this substitution, recall that [tex]\(\csc x = \frac{1}{\sin x}\)[/tex]. Thus, the original equation becomes:
[tex]\[ 4 \sqrt{3} y - 4 - 3 y^2 = 0 \][/tex]
### Step 2: Rearrange the equation
Rearrange the equation into standard quadratic form [tex]\(ay^2 + by + c = 0\)[/tex]:
[tex]\[ -3 y^2 + 4 \sqrt{3} y - 4 = 0 \][/tex]
Multiply through by [tex]\(-1\)[/tex] to simplify:
[tex]\[ 3 y^2 - 4 \sqrt{3} y + 4 = 0 \][/tex]
### Step 3: Use the quadratic formula
The quadratic formula states that for an equation [tex]\(ay^2 + by + c = 0\)[/tex], the solutions for [tex]\(y\)[/tex] are given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 3\)[/tex], [tex]\(b = -4\sqrt{3}\)[/tex], and [tex]\(c = 4\)[/tex]. Plugging these into the quadratic formula, we get:
[tex]\[ y = \frac{-(-4\sqrt{3}) \pm \sqrt{(-4\sqrt{3})^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{4\sqrt{3} \pm \sqrt{48 - 48}}{6} \][/tex]
[tex]\[ y = \frac{4\sqrt{3} \pm \sqrt{0}}{6} \][/tex]
[tex]\[ y = \frac{4\sqrt{3}}{6} \][/tex]
[tex]\[ y = \frac{2\sqrt{3}}{3} \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Recall that [tex]\(y = \csc x = \frac{1}{\sin x}\)[/tex]. So we have:
[tex]\[ \csc x = \frac{2\sqrt{3}}{3} \][/tex]
This implies:
[tex]\[ \sin x = \frac{3}{2\sqrt{3}} \][/tex]
[tex]\[ \sin x = \frac{\sqrt{3}}{2} \][/tex]
### Step 5: Determine [tex]\(x\)[/tex] in the given interval
We need to find [tex]\(x\)[/tex] such that [tex]\(\sin x = \frac{\sqrt{3}}{2}\)[/tex] within the interval [tex]\([0, \frac{\pi}{2}]\)[/tex]. We know from trigonometry that:
[tex]\[ x = \frac{\pi}{3} \][/tex]
### Conclusion
Thus, the solution to the equation [tex]\(4 \sqrt{3} \csc x - 4 - 3 \csc^2 x = 0\)[/tex] within the interval [tex]\([0, \frac{\pi}{2}]\)[/tex] is:
[tex]\[ x = \frac{\pi}{3} \][/tex]
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