How do you find the exact value of $2 {cos}^{2} θ = sin θ + 2$ $2cos^2theta=sintheta+2$ in the interval $0 \leq θ < 360$ $0<=theta<360$ ?

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How do you find the exact value of $2 {cos}^{2} θ = sin θ + 2$ $2cos^2theta=sintheta+2$ in the interval $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2016-12-16T15:30:27

$0; \frac{π}{6}; \frac{5 π}{6}; π; 2 π$ $0; pi/6; (5pi)/6; pi; 2pi$

Explanation:

Replace $2 {cos}^{2} t = 2 - 2 {sin}^{2} t$ $2cos^2 t = 2 - 2sin^2 t$ in the equation:
$2 - 2 {sin}^{2} t = sin t + 2$ $2 - 2sin^2 t = sin t + 2$
$- 2 {sin}^{2} t - sin t = 0$ $- 2sin^2 t - sin t = 0$
sin t( -2sin t + 1) = 0
a. sin t = 0 --> arc $t = 0, t = π, t = 2 π$ $t = 0, t = pi, t = 2pi$
b. - 2sin t + 1 = 0 --> $sin t = \frac{1}{2}$ $sin t = 1/2$
Trig table of special arcs and unit circle give:
$sin t = \frac{1}{2}$ $sin t = 1/2$ --> arc $t = \frac{π}{6}$ $t = pi/6$ and $t = \frac{5 π}{6}$ $t = (5pi)/6$
Answers for (0, 2pi):
$0; \frac{π}{6}; \frac{5 π}{6}; π; 2 π$ $0; pi/6; (5pi)/6; pi; 2pi$

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How do you find the exact value of $2 {cos}^{2} θ = sin θ + 2$ $2cos^2theta=sintheta+2$ in the interval $0 \leq θ < 360$ $0<=theta<360$ ?

1 Answer

Explanation:

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How do you find the exact value of $2 {cos}^{2} θ = sin θ + 2$ $2cos^2theta=sintheta+2$ in the interval $0 \leq θ < 360$ $0<=theta<360$ ?