Solve the following equations.
-\sin^2\left(x\right)+2\cos\left(x\right)=2
For any real number x:
\sin^{2}\left(x\right) = 1 - \cos^{2}\left(x\right)
Therefore:
-\sin^{2}\left(x\right) + 2\cos\left(x\right) = 2
-\left( 1 - \cos^{2}\left(x\right) \right) + 2\cos\left(x\right) = 2
-1 + \cos^{2}\left(x\right) + 2\cos\left(x\right) = 2
\cos^{2}\left(x\right) + 2\cos\left(x\right) - 3 = 0
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) :
- a = 1
- b = 2
- c = -3
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\cos\left(x\right) = \dfrac{-2 \pm \sqrt{2^2 - 4\left(1\right)\left(-3\right)}}{2\left(1\right)}
\cos\left(x\right) = \dfrac{-2 \pm \sqrt{4 + 12}}{2}
\cos\left(x\right) = \dfrac{-2 \pm \sqrt{16}}{2}
\cos\left(x\right) = \dfrac{-2 \pm 4}{2}
\cos\left(x\right) = -1 \pm 2
\begin{cases} \cos\left(x\right) = 1\cr or \cr \cos\left(x\right) = -3 \end{cases}
The \cos\left(x\right) = -3 "solution" is impossible since \cos\left(x\right) has a range of \left[-1{,}1\right]. Therefore, \cos\left(x\right) = 1 is the only real solution.
The cosine function is equal to 1 when its argument is equal to 2n\pi, where n is an integer. Therefore:
x = 2n\pi, \ n \in \mathbb{Z}
x = 2n\pi, n \in \mathbb{Z}
3\cos\left(x\right) + \cos\left(2x\right) = -2
For any real number x:
\cos\left(2x\right) = 2\cos^2\left(x\right) - 1
Therefore:
3\cos\left(x\right) + \cos\left(2x\right) = -2
3\cos\left(x\right) + 2\cos^2\left(x\right) - 1 = -2
2\cos^2\left(x\right) + 3\cos\left(x\right) + 1 = 0
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) :
- a = 2
- b = 3
- c = 1
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\cos\left(x\right) = \dfrac{-3 \pm \sqrt{3^2 - 4\left(2\right)\left(1\right)}}{2\left(2\right)}
\cos\left(x\right) = \dfrac{-3 \pm \sqrt{9 - 8}}{4}
\cos\left(x\right) = \dfrac{-3 \pm \sqrt{1}}{4}
\cos\left(x\right) = \dfrac{-3 \pm 1}{4}
\cos\left(x\right) = \dfrac{-3}{4} \pm \dfrac{1}{4}
\begin{cases} \cos\left(x\right) = \dfrac{-1}{2}\cr or \cr \cos\left(x\right) = -1 \end{cases}
The cosine function is equal to -1 when its argument is equal to \left(2n+1\right)\pi, where n is an integer. Therefore, one set of solutions is:
x = \left(2n+1\right)\pi, n \in \mathbb{Z}
The cosine function is equal to \dfrac{-1}{2} when its argument is equal to \dfrac{4\pi}{3} + 2n\pi or \dfrac{2\pi}{3} + 2n\pi, where n is an integer. Therefore, the rest of the solutions are:
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{2\pi}{3} + 2n\pi, n \in \mathbb{Z}
3\sin\left(2x\right) - \cos\left(x\right)\sin\left(2x\right) = 4\sin\left(x\right)
For any real number x:
\sin\left(2x\right) = 2\sin\left(x\right)\cos\left(x\right)
Therefore:
3\sin\left(2x\right) - \cos\left(x\right)\sin\left(2x\right) = 4\sin\left(x\right)
6\sin\left(x\right)\cos\left(x\right) - 2\sin\left(x\right)\cos^2\left(x\right) = 4\sin\left(x\right)
-3\sin\left(x\right)\cos\left(x\right) + \sin\left(x\right)\cos^2\left(x\right) + 2\sin\left(x\right) = 0
\sin\left(x\right)\left( \cos^2\left(x\right) - 3\cos\left(x\right) + 2 \right) = 0
The solutions to this are the combined solutions to:
\sin\left(x\right) = 0
And:
\cos^2\left(x\right) - 3\cos\left(x\right) + 2 = 0
The sine function is equal to zero when its argument is \pi n, where n is an integer. Therefore, one set of solutions is:
x = \pi n, n \in \mathbb{Z}
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) for \cos^2\left(x\right) - 3\cos\left(x\right) + 2 = 0
- a = 1
- b = -3
- c = 2
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\cos\left(x\right) = \dfrac{3 \pm \sqrt{\left(-3\right)^2 - 4\left(1\right)\left(2\right)}}{2\left(1\right)}
\cos\left(x\right) = \dfrac{3 \pm \sqrt{9 - 8}}{2}
\cos\left(x\right) = \dfrac{3 \pm \sqrt{1}}{2}
\cos\left(x\right) = \dfrac{3 \pm 1}{2}
\cos\left(x\right) = \dfrac{3}{2} \pm \dfrac{1}{2}
\begin{cases} \cos\left(x\right) = 2\cr or \cr \cos\left(x\right) = 1 \end{cases}
The first equation is impossible since \cos\left(x\right) has a range of \left[-1{,}1\right]. Therefore, \cos\left(x\right) = 1 is the only real solution.
The cosine function is equal to 1 when its argument is equal to 2n\pi, where n is an integer. Therefore one set of solutions is:
x = 2n\pi, n \in \mathbb{Z}
This is a subset of the other solution set x = \pi n, n \in \mathbb{Z}.
x = n\pi, n \in \mathbb{Z}
\cos\left(x\right) + \cos\left(2x\right) = -\sin^2\left(x\right)
For any real number x:
\cos\left(2x\right) = \cos^2\left(x\right) - \sin^2\left(x\right)
Therefore:
\cos\left(x\right) + \cos\left(2x\right) = -\sin^2\left(x\right)
\cos\left(x\right) + \cos^2\left(x\right) - \sin^2\left(x\right) = -\sin^2\left(x\right)
\cos^2\left(x\right) + \cos\left(x\right) = 0
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) :
- a = 1
- b = 1
- c = 0
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\cos\left(x\right) = \dfrac{-1 \pm \sqrt{1^2 - 4\left(1\right)\left(0\right)}}{2\left(1\right)}
\cos\left(x\right) = \dfrac{-1 \pm \sqrt{1}}{2}
\cos\left(x\right) = \dfrac{-1 \pm 1}{2}
\cos\left(x\right) = \dfrac{-1}{2} \pm \dfrac{1}{2}
\begin{cases} \cos\left(x\right) = 0\cr or \cr \cos\left(x\right) = -1 \end{cases}
The cosine function is equal to -1 when its argument is equal to \pi \left(2n + 1\right), where n is an integer. Therefore, one set of solutions is:
x = \pi \left(2n + 1\right), n \in \mathbb{Z}
The cosine function is equal to 0 when its argument is equal to \dfrac{\pi}{2}\left(2n + 1\right), where n is an integer.
x = \pi \left(2n + 1\right), n \in \mathbb{Z}\\x = \dfrac{\pi}{2}\left(2n + 1\right), 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)
For any real number x:
\cos\left(2x\right) = \cos^2\left(x\right) - \sin^2\left(x\right)
and
\cos^2\left(x\right) = 1 - \sin^2\left(x\right)
Therefore:
-2\sin^2\left(x\right) + 2 - 4\cos\left(2x\right) + 4\sin\left(x\right) = -2\cos^2\left(x\right)
-2\sin^2\left(x\right) + 2 - 4\left( \cos^2\left(x\right) - \sin^2\left(x\right) \right) + 4\sin\left(x\right) = -2\cos^2\left(x\right)
2\sin^2\left(x\right) + 2 - 4\cos^2\left(x\right) + 4\sin\left(x\right) = -2\cos^2\left(x\right)
\sin^2\left(x\right) + 2\sin\left(x\right) + 1 = \cos^2\left(x\right)
\sin^2\left(x\right) + 2\sin\left(x\right) + 1 = 1 - \sin^2\left(x\right)
2\sin^2\left(x\right) + 2\sin\left(x\right) = 0
\sin^2\left(x\right) + \sin\left(x\right) = 0
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) :
- a = 1
- b = 1
- c = 0
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\sin\left(x\right) = \dfrac{-1 \pm \sqrt{1^2 - 4\left(1\right)\left(0\right)}}{2\left(1\right)}
\sin\left(x\right) = \dfrac{-1 \pm \sqrt{1}}{2}
\sin\left(x\right) = \dfrac{-1 \pm 1}{2}
\sin\left(x\right) = \dfrac{-1}{2} \pm \dfrac{1}{2}
\begin{cases} \sin\left(x\right) = 0\cr or \cr \sin\left(x\right) = -1 \end{cases}
The sine function is equal to 0 when its argument is equal to \pi n, where n is an integer. Therefore, one set of solutions is:
x = \pi n, n \in \mathbb{Z}
The sin function is equal to -1 when its argument is equal to \dfrac{3\pi}{2} + 2\pi n, where n is an integer. Therefore, one set of solutions is:
x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}
x = \pi n, n \in \mathbb{Z}\\x = \dfrac{3\pi}{2} + 2\pi n, n \in \mathbb{Z}
\cos\left(2x\right) + \sin\left(\dfrac{-x}{2}\right)\cos\left(\dfrac{x}{2}\right) = -5
For any real number x:
- \cos\left(2x\right) = \cos^2\left(x\right) - \sin^2\left(x\right)
- \sin\left(2x\right) = 2\sin\left(x\right)\cos\left(x\right)
- \sin\left(-x\right) = -\sin\left(x\right)
- \cos^2\left(x\right) = 1 - \sin^2\left(x\right)
Therefore:
\left( \cos\left(2x\right) \right) + \left( \sin\left(\dfrac{-x}{2}\right)\cos\left(\dfrac{x}{2}\right) \right) = -5
\left( \cos^2\left(x\right) - \sin^2\left(x\right) \right) + \left( -\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right) \right) = -5
\left( \cos^2\left(x\right) - \sin^2\left(x\right) \right) + \left( -\sin\left(x\right) \right) = -5
\left( 1 - \sin^2\left(x\right) - \sin^2\left(x\right) \right) + \left( -\sin\left(x\right) \right) = -5
-2\sin^2\left(x\right) - \sin\left(x\right) + 6 = 0
2\sin^2\left(x\right) + \sin\left(x\right) - 6 = 0
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) :
- a = 2
- b = 1
- c = -6
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\sin\left(x\right) = \dfrac{-1 \pm \sqrt{1^2 - 4\left(2\right)\left(-6\right)}}{2\left(2\right)}
\sin\left(x\right) = \dfrac{-1 \pm \sqrt{1 + 48}}{4}
\sin\left(x\right) = \dfrac{-1 \pm \sqrt{49}}{4}
\sin\left(x\right) = \dfrac{-1 \pm 7}{4}
\sin\left(x\right) = \dfrac{-1}{4} \pm \dfrac{7}{4}
\begin{cases} \sin\left(x\right) = \dfrac{3}{2}\cr or \cr \sin\left(x\right) = -2\end{cases}
Both the \sin\left(x\right) = \dfrac{3}{2} and \sin\left(x\right) = -2 are impossible since \sin\left(x\right) has a range of \left[-1{,}1\right]. Therefore, there are no real solutions.
There are no real solutions.
\tan^2\left(x\right) + \dfrac{3 - \cos\left(2x\right)}{1 + \cos\left(2x\right)} = \cos\left(2x\right)
For any real number x:
\tan^2\left(x\right) = \dfrac{1 - \cos\left(2x\right)}{1 + \cos\left(2x\right)}
Therefore:
\tan^2\left(x\right) + \dfrac{3 - \cos\left(2x\right)}{1 + \cos\left(2x\right)} = \cos\left(2x\right)
\dfrac{1 - \cos\left(2x\right)}{1 + \cos\left(2x\right)} + \dfrac{3 - \cos\left(2x\right)}{1 + \cos\left(2x\right)} = \cos\left(2x\right)
\dfrac{4 - 2\cos\left(2x\right)}{1 + \cos\left(2x\right)} = \cos\left(2x\right)
4 - 2\cos\left(2x\right) = \cos\left(2x\right) + \cos^2\left(2x\right)
\cos^2\left(2x\right) + 3\cos\left(2x\right) - 4 = 0
Use the quadratic equation aX^2+bX+c=0 to find \cos\left(x\right) :
- a = 1
- b = 3
- c = -4
The solutions are X = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Therefore:
\cos\left(2x\right) = \dfrac{-3 \pm \sqrt{3^2 - 4\left(1\right)\left(-4\right)}}{2\left(1\right)}
\cos\left(2x\right) = \dfrac{-3 \pm \sqrt{9 + 16}}{2}
\cos\left(2x\right) = \dfrac{-3 \pm \sqrt{25}}{2}
\cos\left(2x\right) = \dfrac{-3 \pm 5}{2}
\cos\left(2x\right) = \dfrac{-3}{2} \pm \dfrac{5}{2}
\begin{cases} \cos\left(2x\right) = -4\cr or \cr \sin\left(x\right) = 1 \end{cases}
\cos\left(2x\right) = -4 is impossible since \cos\left(2x\right) has a range of \left[-1{,}1\right]. Therefore, \cos\left(2x\right) = 1 is the only real solution.
The cosine function is equal to 1 when its argument is equal to 2\pi n, where n is an integer. Therefore, the solution is:
2x = 2\pi n, n \in \mathbb{Z}
x = \pi n, n \in \mathbb{Z}