Solve s(60^0+θ)*cos(60^0-θ)=1/4cos3θ

$$s ( 60 ^ { 0 } + \theta ) \cdot \cos ( 60 ^ { 0 } - \theta ) = \frac { 1 } { 4 } \cos 3 \theta$$

Solve for s (complex solution)

$$\left\{\begin{matrix}s=\frac{e^{2i\theta +i}+e^{-4i\theta +i}}{4\left(\theta e^{2\left(-i\theta +i\right)}+\theta +e^{2\left(-i\theta +i\right)}+1\right)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}+1\text{ and }4\theta e^{-2i\theta +2i}+4e^{-2i\theta +2i}+4\theta +4\neq 0\\s\in \mathrm{C}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}+1\text{ and }-e^{2i\theta +i}-e^{-4i\theta +i}=0\text{ and }4\theta e^{-2i\theta +2i}+4e^{-2i\theta +2i}+4\theta +4=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 60 to the power of 0 and get 1.

$$s\left(1+\theta \right)\cos(60^{0}-\theta )=\frac{1}{4}\cos(3\theta )$$

Calculate 60 to the power of 0 and get 1.

$$s\left(1+\theta \right)\cos(1-\theta )=\frac{1}{4}\cos(3\theta )$$

Use the distributive property to multiply s by 1+\theta .

$$\left(s+s\theta \right)\cos(1-\theta )=\frac{1}{4}\cos(3\theta )$$

Use the distributive property to multiply s+s\theta by \cos(1-\theta ).

$$s\cos(1-\theta )+s\theta \cos(1-\theta )=\frac{1}{4}\cos(3\theta )$$

Combine all terms containing s.

$$\left(\cos(1-\theta )+\theta \cos(1-\theta )\right)s=\frac{1}{4}\cos(3\theta )$$

The equation is in standard form.

$$\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)s=\frac{\cos(3\theta )}{4}$$

Divide both sides by \cos(1-\theta )+\theta \cos(1-\theta ).

$$\frac{\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)s}{\theta \cos(1-\theta )+\cos(1-\theta )}=\frac{\cos(3\theta )}{4\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)}$$

Dividing by \cos(1-\theta )+\theta \cos(1-\theta ) undoes the multiplication by \cos(1-\theta )+\theta \cos(1-\theta ).

$$s=\frac{\cos(3\theta )}{4\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)}$$

Divide \frac{\cos(3\theta )}{4} by \cos(1-\theta )+\theta \cos(1-\theta ).

$$s=\frac{\cos(3\theta )}{4\left(\theta +1\right)\cos(1-\theta )}$$

Solve for s

$$\left\{\begin{matrix}s=-\frac{\cos(\theta )\left(3\left(\sin(\theta )\right)^{2}-\left(\cos(\theta )\right)^{2}\right)}{4\left(\sin(1)\theta \sin(\theta )+\cos(1)\theta \cos(\theta )+\sin(1)\sin(\theta )+\cos(1)\cos(\theta )\right)}\text{, }&4\left(\sin(1)\theta \sin(\theta )+\cos(1)\theta \cos(\theta )+\sin(1)\sin(\theta )+\cos(1)\cos(\theta )\right)\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\frac{2\pi n_{1}+\pi +2}{2}\\s\in \mathrm{R}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\frac{\pi \left(2n_{2}+1\right)}{6}\text{ and }4\left(\sin(1)\theta \sin(\theta )+\cos(1)\theta \cos(\theta )+\sin(1)\sin(\theta )+\cos(1)\cos(\theta )\right)=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\frac{2\pi n_{1}+\pi +2}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 60 to the power of 0 and get 1.

$$s\left(1+\theta \right)\cos(60^{0}-\theta )=\frac{1}{4}\cos(3\theta )$$

Calculate 60 to the power of 0 and get 1.

$$s\left(1+\theta \right)\cos(1-\theta )=\frac{1}{4}\cos(3\theta )$$

Use the distributive property to multiply s by 1+\theta .

$$\left(s+s\theta \right)\cos(1-\theta )=\frac{1}{4}\cos(3\theta )$$

Use the distributive property to multiply s+s\theta by \cos(1-\theta ).

$$s\cos(1-\theta )+s\theta \cos(1-\theta )=\frac{1}{4}\cos(3\theta )$$

Combine all terms containing s.

$$\left(\cos(1-\theta )+\theta \cos(1-\theta )\right)s=\frac{1}{4}\cos(3\theta )$$

The equation is in standard form.

$$\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)s=\frac{\cos(3\theta )}{4}$$

Divide both sides by \cos(1-\theta )+\theta \cos(1-\theta ).

$$\frac{\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)s}{\theta \cos(1-\theta )+\cos(1-\theta )}=\frac{\cos(3\theta )}{4\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)}$$

Dividing by \cos(1-\theta )+\theta \cos(1-\theta ) undoes the multiplication by \cos(1-\theta )+\theta \cos(1-\theta ).

$$s=\frac{\cos(3\theta )}{4\left(\theta \cos(1-\theta )+\cos(1-\theta )\right)}$$

Divide \frac{\cos(3\theta )}{4} by \cos(1-\theta )+\theta \cos(1-\theta ).

$$s=\frac{\cos(3\theta )}{4\left(\theta +1\right)\cos(1-\theta )}$$