Basic Trig Functions: solve 2cos(3x) + sqrt(3) = 0

[daym]

Hello, really struggling to figure the following question out: "solve 2cos(3x) +sqrt(3)=0. I tried rearranging to get cos(3x) = sqrt(3) / 2 but that didn't give the correct answers. What am I doing wrong? Thanks

 

[stapel]

Hello, really struggling to figure the following question out: "solve 2cos(3x) +sqrt(3)=0. I tried rearranging to get cos(3x) = sqrt(3) / 2 but that didn't give the correct answers. What am I doing wrong?

Since you haven't shown all your work, there is no way to advise you regarding where you might have gone wrong. I can only say that your one posted step is correct, and that this should have led directly to the answer (using the reference-angle values you've memorized).

Kindly please reply showing the rest of your steps, along with what you think are "the correct answers" (the book may have a typo). Thank you!

 

[Ishuda]

Hello, really struggling to figure the following question out: "solve 2cos(3x) +sqrt(3)=0. I tried rearranging to get cos(3x) = sqrt(3) / 2 but that didn't give the correct answers. What am I doing wrong? Thanks

If
2 cos(3x) + $\displaystyle \sqrt{3}$ = 0
then
cos(3x) = - $\displaystyle \frac{\sqrt{3}}{2}$.

 

[Pradeep]

Two Answers

2 cos(3x) + [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT] = 0

cos(3x) = - [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT]

3x = cos^-1(-√3/2)

3x = 150 (or) 210

x = 50^o (or) 70^o (in degrees)

x = 5π/9 (or) 7π/9 (in radians)