If \(1+\sin^2\theta=3\sin\theta\cos\theta\), prove that \(\tan\theta=1\) or \(\dfrac12\).
\(\tan\theta\in\{1,\dfrac12\}.\)
Step 1: We are given:
\[1 + \sin^2\theta = 3\sin\theta \cos\theta.\]
Step 2: To solve this, let us write everything in terms of \(\tan\theta\). Recall the identity: \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}.\)
Step 3: Put \(\tan\theta = t\). Then \(\sin\theta = t \cos\theta.\)
Step 4: Substitute this into the given equation:
\[1 + (t\cos\theta)^2 = 3(t\cos\theta)(\cos\theta).\]
Step 5: Simplify both sides:
\[1 + t^2 \cos^2\theta = 3t \cos^2\theta.\]
Step 6: Divide the whole equation by \(\cos^2\theta\) (valid if \(\cos\theta \neq 0\)):
\[\dfrac{1}{\cos^2\theta} + t^2 = 3t.\]
Step 7: Recall that \(\dfrac{1}{\cos^2\theta} = \sec^2\theta.\) Also, \(\sec^2\theta = 1 + \tan^2\theta = 1 + t^2.\)
Step 8: Substitute this:
\[(1 + t^2) + t^2 = 3t.\]
Step 9: Simplify:
\[1 + 2t^2 = 3t.\]
Step 10: Rearrange into standard quadratic form:
\[2t^2 - 3t + 1 = 0.\]
Step 11: Factorize the quadratic:
\[(2t - 1)(t - 1) = 0.\]
Step 12: Solve for \(t\):
\[t = 1 \quad \text{or} \quad t = \dfrac{1}{2}.\]
Final Answer: Since \(t = \tan\theta\),
\[\tan\theta = 1 \quad \text{or} \quad \tan\theta = \dfrac{1}{2}.\]