Menu Close

Question-195042




Question Number 195042 by sonukgindia last updated on 22/Jul/23
Answered by MM42 last updated on 22/Jul/23
260=2^8 +2^2   ⇒2^(16^(sin^2 x) ) =2^8   &  2^(16^(cos^2 x) ) =2^2   ⇒16^(sin^2 x) =8  &  16^(cos^2 x) =2  ⇒sin^2 x=(3/4)  &  cos^2 x=(1/4)⇒x=kπ±(π/3)  or  ⇒2^(16^(sin^2 x) ) =2^2   &  2^(16^(cos^2 x) ) =2^8   ⇒16^(sin^2 x) =2  &  16^(cos^2 x) =8  ⇒sin^2 x=(1/4)  &  cos^2 x=(3/4)⇒x=kπ±(π/6)
$$\mathrm{260}=\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{16}^{{sin}^{\mathrm{2}} {x}} } =\mathrm{2}^{\mathrm{8}} \:\:\&\:\:\mathrm{2}^{\mathrm{16}^{{cos}^{\mathrm{2}} {x}} } =\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{16}^{{sin}^{\mathrm{2}} {x}} =\mathrm{8}\:\:\&\:\:\mathrm{16}^{{cos}^{\mathrm{2}} {x}} =\mathrm{2} \\ $$$$\Rightarrow{sin}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\&\:\:{cos}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{x}={k}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{16}^{{sin}^{\mathrm{2}} {x}} } =\mathrm{2}^{\mathrm{2}} \:\:\&\:\:\mathrm{2}^{\mathrm{16}^{{cos}^{\mathrm{2}} {x}} } =\mathrm{2}^{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{16}^{{sin}^{\mathrm{2}} {x}} =\mathrm{2}\:\:\&\:\:\mathrm{16}^{{cos}^{\mathrm{2}} {x}} =\mathrm{8} \\ $$$$\Rightarrow{sin}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\&\:\:{cos}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{x}={k}\pi\pm\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *