Find all angles, 0^(@)

Question

Find all angles,  0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree.  6sin^(2)theta-5sin theta+4=3 Answer:  theta=

Bytelearn · Accepted Answer

Simplify Equation: First, we need to simplify the given equation by moving all terms to one side to set the equation to zero. 6sin^(2)theta - 5sin theta + 4 - 3 = 0 6sin^(2)theta - 5sin theta + 1 = 0Factor Quadratic: Next, we will factor the quadratic equation in terms of sin(theta). We are looking for two numbers that multiply to 6*1=6 and add up to -5. The numbers that satisfy these conditions are -3 and -2. So we can write the equation as (3sin(theta) - 1)(2sin(theta) - 1) = 0Solve for sin(theta): Now, we will solve each factor for sin(theta). First, for 3sin(theta) - 1 = 0, we get sin(theta) = 1/3. Second, for 2sin(theta) - 1 = 0, we get sin(theta) = 1/2.Find Angles (1/3): We will find the angles for sin(theta) = 1/3. Since the sine function is positive, the angles must be in the first or second quadrant. Using a calculator, we find theta ≈ arcsin(1/3). Theta ≈ 19.5 degrees or 180 - 19.5 = 160.5 degrees.Find Angles (1/2): Next, we will find the angles for sin(theta) = 1/2. This is a known value on the unit circle, corresponding to theta = 30 degrees and theta = 150 degrees.Final Angles: We have found all possible angles for the given equation within the range 0 <= theta < 360 degrees. The angles are approximately 19.5 degrees, 160.5 degrees, 30 degrees, and 150 degrees.

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $6 \sin ^{2} \theta-5 \sin \theta+4=3$ $\newline$ Answer: $\theta=$

Full solution

More problems from Find the roots of factored polynomials

Related topics

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline6sin⁡2θ−5sin⁡θ+4=3 6 \sin ^{2} \theta-5 \sin \theta+4=3 6sin2θ−5sinθ+4=3\newlineAnswer: θ= \theta= θ=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $6 \sin ^{2} \theta-5 \sin \theta+4=3$ $\newline$ Answer: $\theta=$