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How do you find theta?

Solve for ? sin (theta)=1 sin(θ) = 1 sin ( θ) = 1 Take the inverse sine of both sides of the equation to extract θ θ from inside the sine. θ = arcsin(1) θ = arcsin ( 1) Simplify the right side. Tap for


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prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

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