Question

$\rm Find\: the \: value \: of \: θ$
$\rm \blue{\sin ^2 θ = \sqrt{3} \cos θ}$
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Answer 1

hope it is help full to you have a great day

Answer 2

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ value \: of \: θ \\ \\ given : \\ sin {}^{2} \: θ = \sqrt{3} .cos \: θ \\ \\ we \: know \: that \\ sin {}^{2} θ + cos {}^{2} θ = 1 \\ cos {}^{2} θ + \sqrt{3} .cos \: θ = 1 \\ \\ cos {}^{2} θ + \sqrt{3} .cos \: θ - 1 = 0 \\ comparing \: this \: quadratic \: equation \\ with \: \: a.cos {}^{2} \: θ + b.cos \: θ + c = 0 \\ we \: get \\ a = 1 \\ b = \sqrt{3} \\ c = - 1$

$now \: we \: know \: that \\ Δ = b {}^{2} - 4ac \\ \\ = ( \sqrt{3} \: ) {}^{2} - 4(1)( - 1) \\ \\ = 3 - ( - 4) \\ \\ = 3 + 4 \\ \\ Δ= 7$

$applying \: shreedharacharya \: \\ method \\ we \: get \\ \: \\ cos \: θ = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - \sqrt{3}± \: \sqrt{7} }{(2 \times 1)} \\ \\ we \: have \\ \sqrt{3} = 1.73 \\ \sqrt{7} = 2.64$

$thus \: then \\ cos \: θ = \frac{ - \sqrt{3} + \sqrt{7} }{2} \: \: or \: \: = \frac{ - \sqrt{3} - \sqrt{7} }{2} \\ \\ = \frac{ - 1.73 + 2.64}{2} \: \: or \: \: \frac{ - 1.73 - 2.64}{2} \\ \\ = \frac{0.91}{2} \: \: or \: \: = \frac{ - 4.37}{2} \\ \\ = 0.455 \: \: or \: \: = - 2.185 \\ \\ but \: we \: know \: that \\ f(x) \: of \: cos \: x \\ always \: lie \: in \: range \: of \: \\ - 1 < x < 1 \\ \\ hence \: then \\ \: cos \: θ = - 2.185 \: \: is \: absurd \\ \\ thus \: then \\ cos \: θ = 0.455 \\ \\ θ = cos {}^{ - 1} \: (0.455) \\ \\ further \: you \: can \: use \: logarithms \\ and \: natural \: cosines \: \\ section \: in \: log \: book$