Question

Solve 3cos (3x) - 2 = 0 on the interval [0, 21).13л 25л18' 1818 18π 5π 7π 11π 13π 17π99 9 99No solutionTT 11л 13л 23л 25л 35л18' 18 18 18 18 18

Answer 1

No Solution

Step-by-step explanation:

Given the trigonometric equation:

`√(3)\cos (3x)-2=0,(0,2\pi)`

Add 2 to both sides:

`\begin{gathered} \sqrt[]{3}\cos (3x)-2+2=0+2 \\ \sqrt[]{3}\cos (3x)=2 \end{gathered}`

Divide both sides by √3.

`\begin{gathered} \frac{\sqrt[]{3}\cos (3x)}{\sqrt[]{3}}=\frac{2}{\sqrt[]{3}} \\ \implies\cos (3x)=\frac{2\sqrt[]{3}}{3} \end{gathered}`

Take the arccos of both sides:

`\begin{gathered} 3x=\arccos (\frac{2\sqrt[]{3}}{3}) \\ x=(1)/(3)\arccos (\frac{2\sqrt[]{3}}{3}) \\ x\text{ has no solution for }x\in R \end{gathered}`