anonymous 5 years ago Find all solutions on the interval [0,2pi] 2 cosx sinx - cos x =0 Please show+explain work?

1. anonymous

did you get this yet? if not i will solve.

2. watchmath

3. anonymous

first i would write $2cos(x)sin(x)=cos(x)$ then square to get $4cos^2(x)sin^2(x)=cos^2(x)$ $4cos^2(x)[1-cos^2(x)]=cos^2(x)$ $4cos^2(x)-4cos^4(x)-cos^2(x)=3cos^2(x)-4cos^4(x)=0$

4. anonymous

how am i doing so far?

5. watchmath

If i do this I will move the cos x to the left and factor out the cos x

6. anonymous

$4cos^4(x) - 3cos^2(x)=0$ $cos^2(x)(4cos^2(x)-3)=0$ $cos^2(x)=0$ or $cos^2(x)=\frac{3}{4}$

7. anonymous

$cos(x) = 0$ $x=\frac{\pi}{2}$or $x=-\frac{\pi}{2}$

8. anonymous

proof 2 now that i am on a roll. $f'(x)=3ax^2-4ax-8a$ the zeros are $\frac{4a\pm\sqrt{16a^2+96a}}{6a}=\frac{4a\pm4\sqrt{a^2+6a}}{6a}=$ $\frac{2\pm 2\sqrt{a(a+6)}}{3a}$ and in order for this to be -1 $a(a+6)$ must be a perfect square, a miracle if ever there

9. anonymous

ignore that last gibberish it is wrong and not from here anyway.

10. watchmath

Alternative solution $$\cos x(2\sin x -1)=0$$ $$\cos x=0\Rightarrow x=\pi/2,3\pi/2$$ $$\sin x=1/2\Rightarrow x=\pi/6,5\pi/6$$.

11. anonymous

btw we knew $\pm\frac{\pi}{2}$worked to begin with by inspection.

12. anonymous

right, much swifter and i retire now. yikes.

13. anonymous

Thanks!