8cos^4 x−8cos^2 x+1=0 solution:8cos^2 x(cos^2 x−1)+1=0⇒ −8cos^2 xsin^2 x=−1⇒ sin^2 2x=(1/2)⇒sinx=+_− ((√2)/2)⇒ { ((2x=2kπ+(π/4))),((2x=2kπ+((3π)/4))) :} { ((2x=2kπ−(π/4))),((2x=2kπ+((5π)/4))) :} and so we have { ((x=kπ+(π/8))),((x=kπ+((3π)/8))) :} { ((x=kπ−(π/8))),((x=kπ+((5π)/8))) :} why x=((kπ)/4)+(π/8)? can we solve by another way?


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Question Number 25961 by kaivan.ahmadi last updated on 16/Dec/17
$8cos^4 x−8cos^2 x+1=0 solution:8cos^2 x(cos^2 x−1)+1=0⇒ −8cos^2 xsin^2 x=−1⇒ sin^2 2x=(1/2)⇒sinx=+_− ((√2)/2)⇒ { ((2x=2kπ+(π/4))),((2x=2kπ+((3π)/4))) :} { ((2x=2kπ−(π/4))),((2x=2kπ+((5π)/4))) :} and so we have { ((x=kπ+(π/8))),((x=kπ+((3π)/8))) :} { ((x=kπ−(π/8))),((x=kπ+((5π)/8))) :} why x=((kπ)/4)+(π/8)? can we solve by another way?$
${8 \cos}^{4} x - {8 \cos}^{2} x + 1 = 0$ $solution : {8 \cos}^{2} x (\cos^{2} x - 1) + 1 = 0 \Rightarrow$ $- {8 \cos}^{2} {xsin}^{2} x = - 1 \Rightarrow$ $\sin^{2} 2 x = \frac{1}{2} \Rightarrow sinx = \underline{+} \frac{\sqrt{2}}{2} \Rightarrow$ ${\begin{cases} 2 x = 2 k π + \frac{π}{4} \\ 2 x = 2 k π + \frac{3 π}{4} \end{cases}$ ${\begin{cases} 2 x = 2 k π - \frac{π}{4} \\ 2 x = 2 k π + \frac{5 π}{4} \end{cases}$ $and so we have$ ${\begin{cases} x = k π + \frac{π}{8} \\ x = k π + \frac{3 π}{8} \end{cases}$ ${\begin{cases} x = k π - \frac{π}{8} \\ x = k π + \frac{5 π}{8} \end{cases}$ $why x = \frac{k π}{4} + \frac{π}{8} ?$ $can we solve by another way ?$
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