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Find the set of values of 'a' for which the equation 2^{-1} x

Question

Find the set of values of $^{'} a^{'}$ for which the equation $2 cos^{- 1} x = a + a^{2} (cos^{- 1} x)^{- 1}$ posses a solution.

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James Discussed

Find the set of values of

^{'} a^{'}

for which the equation

2 cos^{- 1} x = a + a^{2} (cos^{- 1} x)^{- 1}

posses a solution.

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

2 cos^{- 1} x = a + a^{2} (cos^{- 1} x)

\frac{2 cos ^{- 1} x}{a} = 1 + \frac{a}{cos ^{- 1} x}

let

\frac{a}{cos ^{- 1} x} = t

x \neq = 1

cos^{- 1} x \neq = a

\frac{2}{t} = 1 + t

t^{2} + t - 2 = 0

(t + 2) (t - 1) = 0

solution are

t = 1 \cup t = - 2

\frac{a}{co s ^{- 1} x} \cup \frac{a}{cos ^{- 1} x} = - 2

a = cos^{- 1} x \cup a = - 2 cos^{- 1} x

a = [0, π] \cup a = [- 2 π, 0]

a \in [- 2 π, 0] \cup (a, π]

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Question Text	Find the set of values of $^{'} a^{'}$ for which the equation $2 cos^{- 1} x = a + a^{2} (cos^{- 1} x)^{- 1}$ posses a solution.
Answer Type	Text solution:1
Upvotes	150