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Question Number 25961 by kaivan.ahmadi last updated on 16/Dec/17

{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 ?