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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. 4cos^(2)theta-9=0 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline4cos⁡2θ−9=0 4 \cos ^{2} \theta-9=0 \newlineAnswer: θ= \theta=

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Csc, sec, and cot of special anglesright chevron icon

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Q. Find all angles, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline4cos⁡2θ−9=0 4 \cos ^{2} \theta-9=0 \newlineAnswer: θ= \theta=
  1. Solve for cos⁡2(θ)\cos^2(\theta): Solve the equation for cos⁡2(θ)\cos^2(\theta).
    4cos⁡2(θ)−9=04\cos^2(\theta) - 9 = 0
    Add 99 to both sides of the equation.
    4cos⁡2(θ)=94\cos^2(\theta) = 9
    Divide both sides by 44.
    cos⁡2(θ)=94\cos^2(\theta) = \frac{9}{4}
    Take the square root of both sides.
    cos⁡(θ)=±94\cos(\theta) = \pm\sqrt{\frac{9}{4}}
    cos⁡(θ)=±32\cos(\theta) = \pm\frac{3}{2}

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