# Math Help - Can someone please tell me where im going wrong

1. ## Can someone please tell me where im going wrong

i keep on getting the wrong answer but cant work out where im going wrong

solve: between 0 and 360

3cos2 = 2cos^2

3(2cos^2 - 1) = 2cos^2

6cos^2 - 3 = 2cos^2

4cos^2 - 3 = 0

cos = sqr 3/4

2. Originally Posted by dno
i keep on getting the wrong answer but cant work out where im going wrong

solve: between 0 and 360

3cos2 = 2cos^2

3(2cos^2 - 1) = 2cos^2

6cos^2 - 3 = 2cos^2

4cos^2 - 3 = 0

cos = sqr 3/4
Where is the unknown? What you've posted is mathemetically meaningless.

I assume you mean $3 \cos (2{\color{red}x}) = 2 \cos^2 {\color{red}x}$.

In which case your last line should be $\cos {\color{red}x} = {\color{red}\pm} \frac{\sqrt{3}}{2}$, that is, $\cos {\color{red}x} = \frac{\sqrt{3}}{2}$ and $\cos x = {\color{red}-} \frac{\sqrt{3}}{2}$.

3. Originally Posted by mr fantastic
Where is the unknown? What you've posted is mathemetically meaningless.

I assume you mean $3 \cos (2{\color{red}x}) = 2 \cos^2 {\color{red}x}$.

In which case your last line should be $\cos {\color{red}x} = {\color{red}\pm} \frac{\sqrt{3}}{2}$, that is, $\cos {\color{red}x} = \frac{\sqrt{3}}{2}$ and $\cos x = {\color{red}-} \frac{\sqrt{3}}{2}$.

couldnt be bothered to right out the fetas and it makes it more confusing

the answers in the back of the book are 30, 150, 210, 330

4. Originally Posted by dno
couldnt be bothered to right out the fetas and it makes it more confusing

the answers in the back of the book are 30, 150, 210, 330
I fixed some typos in my previous reply. Go back and look at the final equations you have to solve. They should be simple to solve.

5. Originally Posted by mr fantastic
I fixed some typos in my previous reply. Go back and look at the final equations you have to solve. They should be simple to solve.
oh yeah i get it now tanks