Leysa 4sin(θ)cos(θ)=2sin(θ)4x

$$4 \sin ( \theta ) \cos ( \theta ) = 2 \sin ( \theta ) 4x$$

Leystu fyrir x

$$\left\{\begin{matrix}\\x=\frac{\cos(\theta )}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}\end{matrix}\right.$$

Skref til að leysa línulega jöfnu

Styttu burt 4 báðum megin.

$$\sin(\theta )\cos(\theta )=2\sin(\theta )x$$

Skipta um hliðar svo allir liðir breytunnar séu vinstra megin.

$$2\sin(\theta )x=\sin(\theta )\cos(\theta )$$

Jafnan er í staðalformi.

$$2\sin(\theta )x=\frac{1}{2}\sin(2\theta )$$

Deildu báðum hliðum með 2\sin(\theta ).

$$\frac{2\sin(\theta )x}{2\sin(\theta )}=\frac{\sin(2\theta )}{2\times 2\sin(\theta )}$$

Að deila með 2\sin(\theta ) afturkallar margföldun með 2\sin(\theta ).

$$x=\frac{\sin(2\theta )}{2\times 2\sin(\theta )}$$

Deildu \frac{1}{2}\sin(2\theta ) með 2\sin(\theta ).

$$x=\frac{\cos(\theta )}{2}$$

Leystu fyrir θ

$$\left\{\begin{matrix}\\\theta =\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\text{unconditionally}\\\theta =-\arccos(2x)+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{; }\theta =\arccos(2x)+2\pi n_{3}\text{, }n_{3}\in \mathrm{Z}\text{, }&|x|\leq \frac{1}{2}\end{matrix}\right.$$