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Sona
please help 2cos(^2)2x=-1+3cos2x
let \(\large y=cos(2x) \) so that trig equation can be written as: \(\large 2y^2=-1+3y \) or \(\large 2y^2-3y+1=0 \) can you solve for y?
so i plug the value for y into one of the equations?
you realize that the trig equation you have is actually a trig equation in quadratic form??? so you would solve the quadratic equation by factoring: \(\large 2y^2-3y+1=0 \) \(\large (2y-1)(y-1)=0 \) so y = ????
but its \[2\cos ^{2}2\theta=-1+3\cos 2\theta\]
n dik wat 2 do w/the 2 in front of the theta
read what i said... the trigonometric equation you have is a trig equation in QUADRATIC form.... to solve that trig equation, you would use the technique used to solve the quadratic equation: \(\large 2y^2=-1+3y \)
so if i use the quadratic formula then i shud b able to solve for y?
ok... this is your equation: \(\large 2\cdot \color {red}{cos^22\theta}=-1+3\cdot \color {red}{cos2\theta} \) \(\large 2\cdot \color {red}{y^2}=-1+3\cdot \color {red}{y} \) sure you can use the quadratic formula but it would be easier to factor as i did earlier...
is y \[(-3\pm sort{17})/4\]
no... rewrite the quadratic in standard form, then factor: \(\large 2y^2=-1+3y \) \(\large 2y^2-3y+1=0 \) standard form \(\large (2y-1)(y-1)=0 \) factor so \(\large 2y-1=0 \) or \(\large y-1=0 \) now it's obvious y=1/2 , y = 1....
now that you have y=1/2 or y=1, remember that we replaced cos22θ=y so now you have to solve these two equations: \(\large cos2\theta=\frac{1}{2} \) ; \(\large cos2\theta=1 \)
you have several answers between [0, 2pi)
or just is it just 30 n 330 since we are dealing with cos?
i only see 30, 330, 0 degrees
well, the 30 and 330 degrees came from the first equation: \(\large cos2\theta=\frac{1}{2} \) and the 0 degrees came from cos2theta = 1
thank u!!!!!!! u helped so much.....by any chance cud u help me solve this next one? 6tanθ−6cotθ=0
yw... post it up as a new question...