91Ó°ÊÓ

Recall the Pythagorean trigonometric identity: $$\tan^2(x) + 1 = \sec^2(x)$$ Using this identity, we can rewrite the given equation as: $$\tan^2(2\theta) = \sec^2(2\theta) - 1$$

Since the equation \(\tan^2(2\theta) = \sec^2(2\theta) - 1\) involves the tangent and secant functions, we can rewrite it in terms of sine and cosine functions, as follows: $$\frac{\sin^2(2\theta)}{\cos^2(2\theta)} = \frac{1}{\cos^2(2\theta)} - 1$$ Now, let's solve for \(\cos^2(2\theta)\): $$\sin^2(2\theta) = 1 - \cos^2(2\theta)$$ In the equation \(\sin^2(2\theta) = 1 - \cos^2(2\theta)\), we use the double-angle formula for sine: $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ So our equation becomes: $$\left(2\sin(\theta)\cos(\theta)\right)^2 = 1 - \cos^2(2\theta)$$ We'll now solve for \(\cos(2\theta)\). To do this, we recall the double-angle formula for cosine: $$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$$ Our equation becomes: $$\left(2\sin(\theta)\cos(\theta)\right)^2 = 1 - \left(\cos^2(\theta) - \sin^2(\theta)\right)^2$$ Since \(\cos^2(\theta) + \sin^2(\theta) = 1\), we have \(\sin^2(\theta) = 1 - \cos^2(\theta)\). Therefore, the equation simplifies to: $$\left(2\sin(\theta)\cos(\theta)\right)^2 = 1 - \left(2\cos^2(\theta) - 1\right)^2$$ Rearranging the equation and factoring, we get: $$(2\cos^2(\theta) - 1)^2 - \left(2\sin(\theta)\cos(\theta)\right)^2 = 0$$ Notice that this is a difference of squares: $$\left[\left(2\cos^2(\theta) - 1\right) + \left(2\sin(\theta)\cos(\theta)\right)\right]\left[\left(2\cos^2(\theta) - 1\right) - \left(2\sin(\theta)\cos(\theta)\right)\right] = 0$$

Now we will solve the two factors separately: Factor 1: $$\left(2\cos^2(\theta) - 1\right) + \left(2\sin(\theta)\cos(\theta)\right) = 0$$ $$2\cos^2(\theta) + 2\sin(\theta)\cos(\theta) - 1 = 0$$ Let \(x = \cos(\theta)\). This gives us the quadratic equation: $$2x^2 + 2x\sqrt{1-x^2} - 1 = 0$$ Solving for \(x\), we obtain \(x = \cos(\theta) = \pm\sqrt{2}/2\). Then, solving for \(\theta\), we find that \(\theta_1 = \pi/4\) and \(\theta_2 = 3\pi/4\) are solutions within the given interval. Factor 2: $$\left(2\cos^2(\theta) - 1\right) - \left(2\sin(\theta)\cos(\theta)\right) = 0$$ $$2\cos^2(\theta) - 2\sin(\theta)\cos(\theta) - 1 = 0$$ Letting \(x = \cos(\theta)\) again, we get another quadratic equation: $$2x^2 - 2x\sqrt{1-x^2} - 1 = 0$$ However, in this case, there are no real solutions for \(x\). Therefore, there are no additional solutions for \(\theta\) in this factor. So the solutions for the given equation are \(\theta=\pi/4\) and \(\theta=3\pi/4\).