The solution of the equation $3 cos h 2 x = 3 + sin h 2 x$ is

0

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\frac{1}{2}

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\frac{1}{2} log 2

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\frac{1}{2} ln 2

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Solution

The correct options are
B $\frac{1}{2}$
D $\frac{1}{2} ln 2$
$3 cos h 2 x - sin h 2 x = 3$
$3 (\frac{e^{2 x} + e^{- 2 x}}{2}) - (\frac{e^{2 x} - e^{- 2 x}}{2}) = 3$
$\Rightarrow 3 e^{2 x} + 3 e^{- 2 x} - e^{2 x} + e^{- 2 x} = 6$
$\Rightarrow 2 e^{2 x} + 4 e^{- 2 x} - 6 = 0$
Multiply the above equation by $e^{2 x}$ we get
$\Rightarrow 2 e^{4 x} - 6 e^{2 x} + 4 = 0$
or $e^{4 x} - 3 e^{2 x} + 2 = 0$
Let $e^{2 x} = α$
$\Rightarrow α^{2} - 3 α + 2 = 0$
$\Rightarrow (α - 2) (α - 1) = 0$
$∴ α = 2, 1$
$\Rightarrow e^{2 x} = 1, 2$
$e^{2 x} = 1$ and $e^{2 x} = 2$
$\Rightarrow e^{2 x} = e^{0} = 1$ and $2 x ln e = ln 2$
$\Rightarrow 2 x = 1$ and $x = \frac{1}{2} ln 2$
$∴ x = \frac{1}{2}$ and $x = \frac{1}{2} ln 2$

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