A solution of the equation ${log}_{2} (sin x + cos x) - {log}_{2} (cos x) + 1 = 0$ in $(- \frac{π}{4}, \frac{π}{4})$ is

{tan}^{- 1} (- \frac{1}{2})

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{tan}^{- 1} (\frac{1}{2})

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\frac{π}{6}

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Solution

The correct option is A ${tan}^{- 1} (- \frac{1}{2})$
${log}_{2} (sin x + cos x) - {log}_{2} (cos x) + 1 = 0 \Rightarrow {log}_{2} (\frac{sin x + cos x}{cos x}) = - 1 \Rightarrow \frac{sin x + cos x}{cos x} = 2^{- 1} \Rightarrow tan x + 1 = \frac{1}{2} \Rightarrow tan x = - \frac{1}{2} \Rightarrow x = {tan}^{- 1} (- \frac{1}{2})$

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Similar questions

4 {cos}^{- 1} x + {sin}^{- 1} x = \frac{π}{2}

x =

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1 : I_{2} > I_{1} > I_{3} > I_{4}

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{log}_{2} (sin x + cos x) - {log}_{2} (cos x) + 1 = 0

(- \frac{π}{4}, \frac{π}{4})

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