If cos {40^o} = x and cos θ = 1 - 2{x^2} , then possible values of θ lying between {0^o} and {360^

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Trigonometrical Equations

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If $\cos {40^o} = x$ and $\cos \theta = 1 - 2{x^2}$, then the possible values of $\theta $ lying between ${0^o}$ and ${360^o}$is

A

${100^o}$ and ${260^o}$

B

${80^o}$ and ${280^o}$

C

${280^o}$ and ${110^o}$

D

${110^o}$ and ${260^o}$

Solution

(a) Here $\cos \theta = 1 – 2{\cos ^2}{40^o}$

$= – (2{\cos ^2}{40^o} – 1)$ $ = – \cos (2 \times {40^o})$

$= – \cos {80^o}$ = $\cos ({180^o} + {80^o}) = \cos ({180^o} – {80^o})$

Hence, $\cos 260^\circ {\rm{and}}\cos 100^\circ $

$i.e.$, $\theta = 100^\circ $ and $260^\circ$.

Standard 11

Mathematics

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