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Select ALL of the correct answers. *


2sin^(4)theta+2sin^(2)theta*cos^(2)theta=1

55. Select ALL of the correct answers. *\newline2sin4θ+2sin2θcos2θ=1 2 \sin ^{4} \theta+2 \sin ^{2} \theta \cdot \cos ^{2} \theta=1

Full solution

Q. 55. Select ALL of the correct answers. *\newline2sin4θ+2sin2θcos2θ=1 2 \sin ^{4} \theta+2 \sin ^{2} \theta \cdot \cos ^{2} \theta=1
  1. Apply Pythagorean identity: Use the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to simplify the equation.\newline2sin4(θ)+2sin2(θ)(1sin2(θ))=12\sin^4(\theta) + 2\sin^2(\theta)(1 - \sin^2(\theta)) = 1
  2. Distribute terms: Distribute the 2sin2θ2\sin^{2}\theta into the parentheses.\newline2sin4θ+2sin2θ2sin4θ=12\sin^{4}\theta + 2\sin^{2}\theta - 2\sin^{4}\theta = 1
  3. Combine like terms: Combine like terms. 2sin4θ2sin4θ+2sin2θ=12\sin^{4}\theta - 2\sin^{4}\theta + 2\sin^{2}\theta = 1
  4. Cancel out terms: The terms 2sin4θ2\sin^{4}\theta cancel each other out.\newline2sin2θ=12\sin^{2}\theta = 1
  5. Isolate sin2(θ)\sin^2(\theta): Divide both sides by 22 to isolate sin2θ\sin^{2}\theta.\newlinesin2θ=12\sin^{2}\theta = \frac{1}{2}
  6. Check solutions: Check if sin2(θ)\sin^2(\theta) can equal 12\frac{1}{2}. This is true for θ=π4\theta = \frac{\pi}{4} or 3π4\frac{3\pi}{4} (in the first and second quadrants where sine is positive).

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