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Question Number 130802 by shaker last updated on 29/Jan/21

	Answered by TheSupreme last updated on 29/Jan/21

	$A s s u m i n g c h = c o s h$ $3 c o s h (2 x) - s i n h (2 x) - 3 = 0$ $3 (\frac{e^{2 x} + e^{- 2 x}}{2}) - \frac{e^{2 x} - e^{- 2 x}}{2} = 3$ $e^{2 x} + 2 e^{- 2 x} - 3 = 0$ $e^{4 x} - 3 e^{2 x} + 2 = 0$ $e^{2 x} = t$ $t^{2} - 3 t + 2 = 0$ $(t - 1) (t - 2) = 0$ $e^{2 x} = 1 \to x = 0$ $e^{2 x} = 2 \to x = \frac{1}{2} \ln (2)$
	Answered by Olaf last updated on 29/Jan/21
	$3ch(2x)−sh(2x)−3 = 0 3(2sh^2 x+1)−2shxchx−3 = 0 2shx(3shx−chx) = 0 ⇒ shx = 0, x = 0 or 3(e^x −e^(−x) )−(e^x +e^(−x) ) = 0 e^x −2e^(−x) = 0 e^(2x) = 2, x = (1/2)ln2 S = {0 ; (1/2)ln2}$
	$3 ch (2 x) - sh (2 x) - 3 = 0$ $3 ({2 sh}^{2} x + 1) - 2 sh x ch x - 3 = 0$ $2 sh x (3 sh x - ch x) = 0$ $\Rightarrow sh x = 0, x = 0$ $or$ $3 (e^{x} - e^{- x}) - (e^{x} + e^{- x}) = 0$ $e^{x} - 2 e^{- x} = 0$ $e^{2 x} = 2, x = \frac{1}{2} \ln 2$ $S = {0; \frac{1}{2} \ln 2}$
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