could someone help me solve this

cos^2x-3/4=0

I know to move the 3/4 to the other side and take the square root, but the answer at the back is 30 and i get 0.5156.

Question

could someone help me solve this 

cos^2x-3/4=0

I know to move the 3/4 to the other side and take the square root, but the answer at the back is 30 and i get 0.5156.

anonymous · Answer

How did you get the 0.5156?

anonymous · Answer

Never mind, I think I figured out how you did it.

anonymous · Answer

You found the square root of 3/4 and rounded it to 0.87 then found the arccosine of that in radian mode.
Put your calculator in degree mode, and take the arccosine of √(3/4) without rounding anything off.

anonymous · Answer

i got 60.4586395 deg. i have to put it in rad, then use the special triangle though.

anonymous · Answer

This can be solved without a calculator using the unit circle.

anonymous · Answer

i know, but there is a way my teachers wants it solved. When i took the inverse cos of 0.87 i got 29.5413605. But in order for me to use the special triangles, i have to round the square root of 3/4 to 0.5 which will be the 30 deg special triangle, but i will be just making the question match what i know it should be

anonymous · Answer

Woah, no, no, no. That will not work.
Do not round any numbers in the middle of calculations.

anonymous · Answer

$$cos(30º)=\sqrt{3/4}$$

anonymous · Answer

i get the answer if i take the whole number, but on a test how will i know if i should take the whole number or just round it. Because we always round to two or 6 decimal places depending on what we are doing.

anonymous · Answer

Never round if you don't have to. If you are using a calculator to approximate you'll want to keep at least 4-5 digits after the decimal point for trig functions. You're better off doing all calculations with exact values or keeping as many digits of precision as possible and only ever round off a number at the very end for your final answer.

anonymous · Answer

For any of the special right triangle angles and ratios, only use exact values.

anonymous · Answer

i know, but the answer is not exact, when i take the square root i get 0.87, which is not any of the special trangles

anonymous · Answer

triangles*

anonymous · Answer

When you take the square root you don't get 0.87 - that's what I'm trying to tell you. Don't round off the numbers. The square root of 3/4 is
 $\large \frac{\sqrt{3}}{2}$

 $\large cos^{-1}(\frac{\sqrt{3}}{2})=30 º$

anonymous · Answer

omg, i should have noticed that. I kept on thinking i was using sin. I'm so stupid. 

Thank you so much