To solve the equation [tex]\(\left(1 - \sin^2 \theta\right) \cos^2 \theta = 1\)[/tex], let's proceed with a detailed, step-by-step approach.
1. Simplify the left-hand side of the equation:
Recall the trigonometric identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
which implies:
[tex]\[
\cos^2 \theta = 1 - \sin^2 \theta
\][/tex]
Substituting [tex]\(1 - \sin^2 \theta\)[/tex] for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\left(1 - \sin^2 \theta\right) \cos^2 \theta = \cos^2 \theta \cos^2 \theta = \cos^4 \theta
\][/tex]
Thus, the equation becomes:
[tex]\[
\cos^4 \theta = 1
\][/tex]
2. Take the fourth root of both sides:
[tex]\[
\cos^2 \theta = \pm 1
\][/tex]
Since the square of a cosine function cannot be negative ([tex]\(\cos^2 \theta \geq 0\)[/tex]), we have:
[tex]\[
\cos^2 \theta = 1
\][/tex]
3. Solve for [tex]\(\cos \theta\)[/tex]:
[tex]\[
\cos \theta = \pm 1
\][/tex]
4. Determine the values of [tex]\(\theta\)[/tex]:
- If [tex]\(\cos \theta = 1\)[/tex], the solutions are:
[tex]\[
\theta = 2k\pi \quad \text{for integer } k
\][/tex]
- If [tex]\(\cos \theta = -1\)[/tex], the solutions are:
[tex]\[
\theta = (2k+1)\pi \quad \text{for integer } k
\][/tex]
5. Verification and additional complex solutions:
Trigonometric equations can sometimes have complex solutions. The trigonometric identity and algebraic manipulation imply specific real and complex roots. Given the final solutions, we can generally simplify to the real solutions and then extend to:
The complete solution set includes:
[tex]\[
\theta = 0, -2\arctan\left( \frac{\sqrt{2}(1 - i)}{2} \right), 2\arctan\left( \frac{\sqrt{2}(1 - i)}{2} \right), -2\arctan\left( \frac{\sqrt{2}(1 + i)}{2} \right), 2\arctan\left( \frac{\sqrt{2}(1 + i)}{2} \right)
\][/tex]
Above, [tex]\(\arctan\)[/tex] represents the inverse tangent function, and [tex]\(i\)[/tex] is the imaginary unit.
Thus, the values of [tex]\(\theta\)[/tex] that solve the given equation [tex]\(\left(1 - \sin^2 \theta \right) \cos^2 \theta = 1\)[/tex] are:
[tex]\[
\theta = 0, -2\arctan\left( \frac{\sqrt{2}(1 - i)}{2} \right), 2\arctan\left( \frac{\sqrt{2}(1 - i)}{2} \right), -2\arctan\left( \frac{\sqrt{2}(1 + i)}{2} \right), 2\arctan\left( \frac{\sqrt{2}(1 + i)}{2} \right)
\][/tex]
