Trig Equations with Multiple Trig Functions cont.

**~berserk** · April 6th 2008, 05:23 PM

Find, to the nearest tenth of a degree, all values of x in the interval 0≤x≤360 that satisfy the equation.

tan x=cos x

also solve for x in the interval 0≤x≤2π

cos ½x= cos x

**janvdl** · April 7th 2008, 01:39 AM

Originally Posted by ~berserk

tan x=cos x

$\frac{sin x}{cos x} = cos x$

$sin x = cos^2 x$

$sin x = 1 - sin^2 x$

$sin^2 x + sin x - 1 = 0$

Solve for x. (Let Sin x = k and solve using the quadratic formula.)

Originally Posted by ~berserk

cos ½x= cos x

$cos x - cos \frac{x}{2} = 0$

This is true where x is 0 and there also seems to be a solution between 45 and 60 degrees, as well as between 15 and 30 degrees. I suggest you draw an accurate graph to find the points.

**Mathstud28** · April 7th 2008, 08:41 AM

$cos\bigg(\frac{x}{2}\bigg)=cos(x)$ ...now using the identity for $cos\bigg(\frac{x}{2}\bigg)$ you obtain $\sqrt{\frac{1+cos(x)}{2}}=cos(x)$ ...now by squaring each side and simplifying you get $2cos(x)^2-1-cos(x)=0$ Define u $sin(x)$ and solve by quadratic formula

**~berserk** · April 7th 2008, 05:43 PM

i think i messed up the second problem yesterday because I did it and solved it like this

**Mathstud28** · April 7th 2008, 05:50 PM

Originally Posted by ~berserk

i think i messed up the second problem yesterday because I did it and solved it like this

IDk...but here you go you factored it right so therefore $cos(x)=1$ ...so $x=cos^{-1}(1)=0$ .... $cos\bigg(\frac{1}{2{\cdot{0}/bigg)=cos(0)$ ...and doing the other one you get $x=\frac{2\pi}{3}$ which is a valid solution

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Trig Equations with Multiple Trig Functions cont.

for the second one I suggest doing it this way

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