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ByteMe Group TitleBest ResponseYou've already chosen the best response.0
let \(\large y=cos(2x) \) so that trig equation can be written as: \(\large 2y^2=1+3y \) or \(\large 2y^23y+1=0 \) can you solve for y?
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
so i plug the value for y into one of the equations?
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
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: \(\large 2y^23y+1=0 \) \(\large (2y1)(y1)=0 \) so y = ????
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
but its \[2\cos ^{2}2\theta=1+3\cos 2\theta\]
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
n dik wat 2 do w/the 2 in front of the theta
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
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: \(\large 2y^2=1+3y \)
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
so if i use the quadratic formula then i shud b able to solve for y?
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
is y \[(3\pm sort{17})/4\]
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
no... rewrite the quadratic in standard form, then factor: \(\large 2y^2=1+3y \) \(\large 2y^23y+1=0 \) standard form \(\large (2y1)(y1)=0 \) factor so \(\large 2y1=0 \) or \(\large y1=0 \) now it's obvious y=1/2 , y = 1....
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
n y stands for cos2theta?
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
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: \(\large cos2\theta=\frac{1}{2} \) ; \(\large cos2\theta=1 \)
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
yes... y=cos2theta...
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
you have several answers between [0, 2pi)
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
so 30 150 210 n 330?
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
or just is it just 30 n 330 since we are dealing with cos?
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
i only see 30, 330, 0 degrees
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
thank u!!!!!!! u helped so much.....by any chance cud u help me solve this next one? 6tanθ−6cotθ=0
 one year ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
yw... post it up as a new question...
 one year ago

Sona Group TitleBest ResponseYou've already chosen the best response.0
okie dokie :)
 one year ago
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