How do you find the exact value of $4 {cos}^{2} θ + 4 sin θ - 5 = 0$ $4cos^2theta+4sintheta-5=0$ in the interval $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2016-12-17T14:27:26

$θ = 30^{o}$ $theta=30^o$ or $150^{o}$ $150^o$

$4 {cos}^{2} θ + 4 sin θ - 5 = 0$ $4cos^2theta+4sintheta-5=0$

$\Leftrightarrow 4 (1 - {sin}^{2} θ) + 4 sin θ - 5 = 0$ $hArr4(1-sin^2theta)+4sintheta-5=0$

or $- 4 {sin}^{2} θ + 4 sin θ - 1 = 0$ $-4sin^2theta+4sintheta-1=0$

or $4 {sin}^{2} θ - 4 sin θ + 1 = 0$ $4sin^2theta-4sintheta+1=0$

or ${(2 sin θ - 1)}^{2} = 0$ $(2sintheta-1)^2=0$

i.e. $2 sin θ - 1 = 0$ $2sintheta-1=0$ or $sin θ = \frac{1}{2} = {sin 30}^{o}$ $sintheta=1/2=sin30^o$ or ${sin 150}^{o}$ $sin150^o$

and $θ = 30^{o}$ $theta=30^o$ or $150^{o}$ $150^o$

How do you find the exact value of 4cos^2theta+4sintheta-5=04cos2θ+4sinθ−5=04cos^2theta+4sintheta-5=0 in the interval 0<=theta<3600≤θ<3600<=theta<360?