Solve quadratric equations that involve trigonometric functions - High school Algebra II Exercises

Solve the following equations.

-\sin^2\left(x\right)+2\cos\left(x\right)=2

x = 2n\pi,\ n \in \mathbb{Z}

x = n\pi, \ n \in \mathbb{Z}

x = \left(2n + 1\right)\pi, \ n \in \mathbb{Z}

x = \dfrac{\pi}{2}\left(2n + 1\right),\ n \in \mathbb{Z}

3\cos\left(x\right) + \cos\left(2x\right) = -2

x = \left(2n+1\right)\pi, n \in \mathbb{Z}

x = \dfrac{4\pi}{3} + 2n\pi, n \in \mathbb{Z}

x = \dfrac{2\pi}{3} + 2n\pi, n \in \mathbb{Z}

x = n\pi, n \in \mathbb{Z}

x = \dfrac{4\pi}{3} + 2n\pi, n \in \mathbb{Z}

x = \dfrac{2\pi}{3} + 2n\pi, n \in \mathbb{Z}

x = \left(2n+1\right)\pi, n \in \mathbb{Z}

x = \dfrac{4\pi}{3} + 2n\pi, n \in \mathbb{Z}

x = \dfrac{-\pi}{3} + 2n\pi, n \in \mathbb{Z}

The equation has no real solutions.

3\sin\left(2x\right) - \cos\left(x\right)\sin\left(2x\right) = 4\sin\left(x\right)

x = \pi n, n \in \mathbb{Z}

x = \pi n, n \in \mathbb{Z}\\x = \dfrac{\pm \pi}{3} + 2\pi n, n \in \mathbb{Z}

x = \pi \left(2n + 1\right), n \in \mathbb{Z}\\x = \dfrac{\pi}{3} + 2\pi n, n \in \mathbb{Z}\\x = \dfrac{5\pi}{3} + 2\pi n, n \in \mathbb{Z}

The equation has no real solutions.

\cos\left(x\right) + \cos\left(2x\right) = -\sin^2\left(x\right)

x = \pi \left(2n + 1\right), n \in \mathbb{Z}\\x = \dfrac{\pi}{2}\left(2n + 1\right), n \in \mathbb{Z}

x = 2n\pi, n \in \mathbb{Z}\\x = \dfrac{\pi}{2}\left(2n + 1\right), n \in \mathbb{Z}

x = n\pi, n \in \mathbb{Z}\\x = \dfrac{\pi n}{2}, n \in \mathbb{Z}

x = 2n\pi, n \in \mathbb{Z}\\x = \dfrac{\pm \pi}{3} + 2\pi, n \in \mathbb{Z}

-2\sin^2\left(x\right) + 2 - 4\cos\left(2x\right) + 4\sin\left(x\right) = -2\cos^2\left(x\right)

x = \pi n, n \in \mathbb{Z}\\x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}

x = 2 \pi n, n \in \mathbb{Z}\\x = \dfrac{1\pi}{2} + 2\pi n, n \in \mathbb{Z}

x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}

x = \pi n, n \in \mathbb{Z}

\cos\left(2x\right) + \sin\left(\dfrac{-x}{2}\right)\cos\left(\dfrac{x}{2}\right) = -5

There are no real solutions.

x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}

x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}\\x = \dfrac{\pi}{6} + 2\pi n, n \in \mathbb{Z}

x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}\\x = \dfrac{\pi}{6} + 2\pi n, n \in \mathbb{Z}\\x = \dfrac{5\pi}{6} + 2\pi n, n \in \mathbb{Z}

\tan^2\left(x\right) + \dfrac{3 - \cos\left(2x\right)}{1 + \cos\left(2x\right)} = \cos\left(2x\right)

x = \pi n, n \in \mathbb{Z}

x = 2\pi n, n \in \mathbb{Z}

x = \dfrac{\pi}{2} n, n \in \mathbb{Z}

There are no real solutions.