Question

8sin⁴ ½theta-8 sin² ½ theta +1 = cos 2 theta​

Answer 1

$8 {sin}^{4} ( \frac{x}{2} ) - 8 {sin}^{2} ( \frac{x}{2} ) + 1 = cos(2x) \\ - 4 {sin}^{2} ( \frac{x}{2} )( 1 + 1 - 2{sin}^{2} ( \frac{x}{2} ) ) + 1 = \cos(2x) \\ - 4 {sin}^{2} ( \frac{x}{2} )(1 + \cos(x) ) + 1 = 1 - 2 {sin}^{2} ( \frac{x}{2} ) \\ - 4 {sin}^{2} ( \frac{x}{2} )(1 + \cos(x) ) = - 2 {sin}^{2} ( \frac{x}{2} ) \\ 2(1 + \cos(x) ) = 1 \\ \cos(x) = - \frac{1}{2} \\ x = \frac{2\pi}{3} = 120°$

Extra information:

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered}$