If \(2\sin^2\theta-\cos^2\theta=2\), find \(\theta\).
\(\theta=90^\circ\).
Step 1: Recall the basic trigonometric identity:
\(\sin^2\theta + \cos^2\theta = 1\).
This means \(\cos^2\theta = 1 - \sin^2\theta\).
Step 2: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) into the given equation:
\(2\sin^2\theta - (1 - \sin^2\theta) = 2\).
Step 3: Simplify the equation:
\(2\sin^2\theta - 1 + \sin^2\theta = 2\).
\(3\sin^2\theta - 1 = 2\).
Step 4: Add 1 to both sides:
\(3\sin^2\theta = 3\).
Step 5: Divide both sides by 3:
\(\sin^2\theta = 1\).
Step 6: Take square root on both sides:
\(\sin\theta = \pm 1\).
Step 7: Now check where sine is equal to 1 or -1:
Step 8: Since we are usually asked for the acute principal angle, the required solution is:
\(\theta = 90^\circ\).