Solve the following equations.
\cos^2\left(x\right)=\sin^2\left(x\right)
For any real number x:
\cos^{2}\left(x\right) - \sin^{2}\left(x\right) = \cos\left(2x\right)
Therefore:
\cos^{2}\left(x\right) = \sin^{2}\left(x\right) :
\cos^{2}\left(x\right) - \sin^{2}\left(x\right) = 0
\cos\left(2x\right) = 0
The cosine function is equal to zero at intervals of \dfrac{\pi}{2}\left(2n + 1\right), where n is an integer \left(\dfrac{-3\pi}{2}, \dfrac{-\pi}{2}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}, ...\right).
Therefore:
2x = \dfrac{\pi}{2}\left(2n + 1\right)
x = \dfrac{\pi}{4}\left(2n + 1\right)
x = \dfrac{\pi}{4}\left(2n + 1\right),\ n \in \mathbb{Z}
\sin\left(2x\right)+\sin^{2}\left(x\right)=\cos^{2}\left(x\right)
Rearrange the equation to get:
\sin\left(2x\right)=\cos^{2}\left(x\right)-\sin^{2}\left(x\right)
Since \cos^{2}\left(\theta\right)-\sin^{2}\left(\theta\right)=\cos\left(2\theta\right) :
\sin\left(2x\right)=\cos\left(2x\right)
Dividing both sides by \cos\left(2x\right) :
\dfrac{\sin\left(2x\right)}{\cos\left(2x\right)}=\dfrac{\cos\left(2x\right)}{\cos\left(2x\right)}
Applying identity \tan\left(\theta\right)=\dfrac{\sin\left(\theta\right)}{\cos\left(\theta\right)} :
\tan\left(2x\right)=1
The tangent function is equal to 1 at intervals of \left(\dfrac{\pi}{4}+n\pi \right), where n is an integer. Therefore:
2x=\dfrac{\pi}{4}+n\pi
Dividing both sides by 2:
x=\dfrac{\pi}{8}+\dfrac{n\pi}{2}
\sqrt{3}\cos\left(x\right)+\sin\left(x\right)=2
Divide both sides by 2:
\dfrac{\sqrt{3}\cos\left(x\right)+\sin\left(x\right)}{2}=\dfrac{2}{2}
\dfrac{\sqrt{3}}{2}\cos\left(x\right)+\dfrac{1}{2}\sin\left(x\right)=1
Applying \sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2} and \cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2} :
\sin\left(\dfrac{\pi}{3}\right)\cos\left(x\right)+\cos\left(\dfrac{\pi}{3}\right)\sin\left(x\right)=1
Applying \sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\cos\left(a\right)\sin\left(b\right) :
Here:
- a=\dfrac{\pi}{3}
- b=x
Therefore:
\sin\left(\dfrac{\pi}{3}+x\right)=1
The sine function is equal to 1 at intervals of \left(\dfrac{\pi}{2}+{2}n\pi \right), where n is an integer. Therefore:
\dfrac{\pi}{3}+x=\dfrac{\pi}{2}+{2}n\pi
The solution is:
x=\dfrac{\pi}{6}+{2}n\pi,\ n \in \mathbb{Z}
\cos\left(x\right)\cos\left(\dfrac{\pi}{5}\right)=\sin\left(x\right)\sin\left(\dfrac{\pi}{5}\right)
\cos\left(x\right)\cos\left(\dfrac{\pi}{5}\right)=\sin\left(x\right)\sin\left(\dfrac{\pi}{5}\right)
\cos\left(x\right)\cos\left(\dfrac{\pi}{5}\right)-\sin\left(x\right)\sin\left(\dfrac{\pi}{5}\right)=0
Applying the indentity \cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right) :
- a=x
- b=\dfrac{\pi}{5}
\cos\left(x+\dfrac{\pi}{5}\right)=0
The cosine function is equal to zero at intervals of \dfrac{\pi}{2}\left(2n + 1\right), where n is an integer. Therefore:
x+\dfrac{\pi}{5}=\dfrac{\pi}{2}\left(2n + 1\right)
The solution is:
x=\dfrac{3\pi}{10}+n\pi, \ n \in \mathbb{Z}
\sin\left(x-\dfrac{\pi}{4}\right)=\sqrt{3}\cos\left(x-\dfrac{\pi}{4}\right)
Dividing both sides by \cos\left(x-\dfrac{\pi}{4}\right) :
\dfrac{\sin\left(x-\dfrac{\pi}{4}\right)}{\cos\left(x-\dfrac{\pi}{4}\right)}=\dfrac{\sqrt{3}\cos\left(x-\dfrac{\pi}{4}\right)}{\cos\left(x-\dfrac{\pi}{4}\right)}
\dfrac{\sin\left(x-\dfrac{\pi}{4}\right)}{\cos\left(x-\dfrac{\pi}{4}\right)}=\sqrt{3}
Applying the identity \tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)} :
\tan\left(x-\dfrac{\pi}{4}\right)=\sqrt{3}
The tangent function is equal to \sqrt{3} at intervals of \left(\dfrac{\pi}{3}+n\pi \right), where n is an integer.
Therefore:
x-\dfrac{\pi}{4}=\dfrac{\pi}{3}+n\pi
The solution is:
x=\dfrac{7\pi}{12}+n\pi,\ n \in \mathbb{Z}
\dfrac{1-\tan^{2}\left(x\right)}{1+\tan^{2}\left(x\right)}=1
Applying the identity \cos\left(2x\right)=\dfrac{1-\tan^{2}\left(x\right)}{1+\tan^{2}\left(x\right)} :
\cos\left(2x\right)=1
The cosine function is equal to 1 at intervals of 2n\pi, where n is an integer. Therefore:
2x=2n\pi
The solution is:
x=n\pi
\sin^{2}\left(x\right)+\cos^{2}\left(x\right)-2\sin\left(x\right)+1=0
Applying the identity \sin^{2}\left(x\right)+\cos^{2}\left(x\right)=1 :
1-2\sin\left(x\right)+1=0
2-2\sin\left(x\right)=0
-2\sin\left(x\right)=-2
\sin\left(x\right)=1
The sine function is equal to 1 at intervals of \left(\dfrac{\pi}{2}+{2}n\pi \right), where n is an integer. Therefore:
x=\dfrac{\pi}{2}+{2}n\pi
x=\dfrac{\pi}{2}+{2}n\pi, \ n \in \mathbb{Z}