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solve 3cosx = 2+cosx​

Sagot :

Sprntidn

Answer:

x= 0.

Step-by-step explanation:

1. Write the expression.

[tex]3cosx = 2+cosx[/tex]

2. Divide both sides by 3.

[tex]\frac{3cosx}{3} = \frac{2+cosx}{3} \\\\\frac{3cosx}{3} = \frac{2}{3} +\frac{cosx}{3} \\ \\cosx=\frac{2}{3} +\frac{cosx}{3}[/tex]

3. Subtract [tex]\frac{cosx}{3}[/tex] from both sides.

[tex]\frac{3cosx}{3} = \frac{2+cosx}{3} \\\\\frac{3cosx}{3} = \frac{2}{3} +\frac{cosx}{3} \\ \\cosx-\frac{cosx}{3}=\frac{2}{3} +\frac{cosx}{3}-\frac{cosx}{3}\\ \\\\cosx-\frac{cosx}{3}=\frac{2}{3}[/tex]

4. Simplify the fraction subtraction.

[tex]\frac{3cosx}{3} -\frac{cosx}{3}=\frac{2}{3}\\ \\\frac{3cosx-cosx}{3}=\frac{2}{3}[/tex]

5. Multiply both sides by 3.

[tex]3*\frac{3cosx-cosx}{3}=\frac{2}{3}*3\\ \\3cosx-cosx=\frac{6}{3}\\ \\3cosx-cosx=2[/tex]

6. Solve the subtraction.

[tex]2cosx=2[/tex]

7. Divide both sides by 2.

[tex]\frac{2cosx}{2} =\frac{2}{2} \\ \\cosx=1[/tex]

8. Take [tex]cos^{-1}[/tex] from both sides.

[tex]cos^{-1}(cosx) =cos^{-1}(1)\\ \\x=cos^{-1}(1)\\ \\x=0.[/tex]

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