Let's solve the equation [tex]\(\frac{3 \cos ^2 2x}{1 + \sin 2x} = 1\)[/tex] within the interval [tex]\(0^{\circ} \leqslant x \leqslant 90^{\circ}\)[/tex].
1. Rewrite the Equation:
We start with the given equation:
[tex]\[
\frac{3 \cos^2 2x}{1 + \sin 2x} = 1
\][/tex]
2. Isolate the Trigonometric Terms:
Multiply both sides by [tex]\((1 + \sin 2x)\)[/tex] to rid of the denominator:
[tex]\[
3 \cos^2 2x = 1 + \sin 2x
\][/tex]
3. Rearrange the Equation:
Move all terms to one side to facilitate solving:
[tex]\[
3 \cos^2 2x - \sin 2x - 1 = 0
\][/tex]
4. Solve for [tex]\(2x\)[/tex]:
Let [tex]\(u = 2x\)[/tex]. The equation becomes easier to handle:
[tex]\[
3 \cos^2 u - \sin u - 1 = 0
\][/tex]
5. Use Trigonometric Identities:
Recall the identity [tex]\(\cos^2 u = 1 - \sin^2 u\)[/tex]. Substitute it into the equation:
[tex]\[
3 (1 - \sin^2 u) - \sin u - 1 = 0
\][/tex]
Expanding and simplifying, we get:
[tex]\[
3 - 3 \sin^2 u - \sin u - 1 = 0
\][/tex]
[tex]\[
-3 \sin^2 u - \sin u + 2 = 0
\][/tex]
Multiply the entire equation by -1 to simplify:
[tex]\[
3 \sin^2 u + \sin u - 2 = 0
\][/tex]
6. Solve the Quadratic Equation:
This is a quadratic equation in terms of [tex]\(\sin u\)[/tex]. Use the quadratic formula [tex]\(a \sin^2 u + b \sin u + c = 0\)[/tex] where [tex]\(a = 3\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -2\)[/tex]:
[tex]\[
\sin u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substituting [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\sin u = \frac{-1 \pm \sqrt{1 + 24}}{6}
\][/tex]
[tex]\[
\sin u = \frac{-1 \pm 5}{6}
\][/tex]
This gives two solutions:
[tex]\[
\sin u = \frac{4}{6} = \frac{2}{3} \quad \text{and} \quad \sin u = \frac{-6}{6} = -1
\][/tex]
7. Convert Back to [tex]\(x\)[/tex]:
Recall that [tex]\(u = 2x\)[/tex]. Thus,
[tex]\[
\sin 2x = \frac{2}{3}
\][/tex]
Check if [tex]\(\sin 2x = -1\)[/tex]:
[tex]\[
2x = \sin^{-1}(-1)
\][/tex]
This solution, however, falls outside the given range [tex]\(0^\circ \leq x \leq 90^\circ\)[/tex].
8. Find [tex]\(x\)[/tex]:
Solve for [tex]\(x\)[/tex] in the valid range:
[tex]\[
2x = \sin^{-1} \left(\frac{2}{3}\right)
\][/tex]
Divide by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{\sin^{-1} \left(\frac{2}{3}\right)}{2}
\][/tex]
[tex]\[
x \approx 0.364863828113483 \quad \text{radians}
\][/tex]
9. Convert to Degrees:
Since [tex]\(x\)[/tex] must be within [tex]\(0^\circ \leq x \leq 90^\circ\)[/tex], we also find another solution within this interval depending on the periodicity and symmetry of the sine function. The numerical results in radians approximate to:
[tex]\[
x_1 = 0.364863828113483 \, \text{radians} \approx 20.9^\circ
\][/tex]
[tex]\[
x_2 = 1.20593249868141 \, \text{radians} \approx 69.1^\circ
\][/tex]
Thus, the solutions to the equation [tex]\(\frac{3 \cos ^2 2x}{1+\sin 2x}=1\)[/tex] within the range [tex]\(0^{\circ} \leqslant x \leqslant 90^{\circ}\)[/tex] are [tex]\( x \approx 20.9^\circ \)[/tex] and [tex]\( x \approx 69.1^\circ \)[/tex].
