sin^2(2x)-cos(2x)-1=0

asked by guest
on Jan 11, 2025 at 8:02 am



You asked:

Solve the equation \({\sin\left( 2 x \right)}^{2} - \cos\left( 2 x \right) - 1 = 0\) for the variable \(x\).

MathBot Answer:

The real solutions are: \[\begin{aligned}x &= \left\{\frac{\pi \left(1 + 4 n\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.5707963\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(3 + 4 n\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(5 + 8 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.9269908\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(3 + 8 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 2.3561945\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(7 + 8 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 5.4977871\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(1 + 8 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.78539816\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

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