Are My Trigonometric Equation Solutions Correct?

[sarah786]

solve the trigonometric equations
tan 2(theta) + cot(theta) = 0

When i solved this question, i got pi/2 + 2n*pi and 3*pi/2 + 2n*pi ... but my book says the answer should be pi/6 + n*pi and 5pi/6 + n*pi

Is my answer wrong?? if so, then what's your solution.?

 

[tiny-tim]

hi sarah786!

(have a pi: π and a theta: θ )

your solution is correct

i think the question was intended to be tan(2θ) minus cotθ = 0

 

[sarah786]

@ tiny-tim.. thank you! :)

 

[sarah786]

can you please correct me with some other questions?? i have solved them but my answers don't match the book's...
√3 tan x - sec x - 1 = 0 ... i get three solutions to this question but my book's got two... it doesn't mention 5pi/3 + 2n*pi... do u think this is also a soln to this question??

sin2x + sin x = 0 ... one of the solutions that i get is 4pi/3 + 2n*pi ... my book doesn't mention this... am i correct??

sin 4x - sin 2x = cos 3x ... one of the solutions that i get is pi/2 + 2n*pi/3 ... am i right?

sin x + sin 3x + sin 5x = 0 ... are pi/3 + 2n*pi and 2n*pi/3 also solutions to this eq. ??

LAST ONE :) ... sin θ + sin 3θ + sin 5θ + sin 7θ = 0 ... is 3*pi/4 + n*pi also a solution...

I really need to confirm the answers i am having my grand test tomorrow... i have tried them but not sure about the answers... i will be grateful forever if you confirm these... thanks :)

 

[tiny-tim]

hi sarah786!

(what happened to that π i gave you? )

√3 tan x - sec x - 1 = 0 ... i get three solutions to this question but my book's got two... it doesn't mention 5pi/3 + 2n*pi... do u think this is also a soln to this question??

let's see …

tan(5π/3) = tan(-π/3) = -tan(π/3) = -tan(60°) = -√3

sec(5π/3) = sec(-π/3) = sec(π/3) = sec(60°) = 2

-3 -2 - 1 = -6, no not a solution …

how did you get it? ​

sin2x + sin x = 0 ... one of the solutions that i get is 4pi/3 + 2n*pi ... my book doesn't mention this... am i correct??

sin(8π/3) = sin(2π/3) = sin120° = sin60° = √3/2

sin(4π/3) = sin(-2π/3) = -sin(2π/3) = -√3/2,

yes that's a solution …

again, how did you get it?​

 

[Mark44]

can you please correct me with some other questions?? i have solved them but my answers don't match the book's...
√3 tan x - sec x - 1 = 0 ... i get three solutions to this question but my book's got two... it doesn't mention 5pi/3 + 2n*pi... do u think this is also a soln to this question??

sin2x + sin x = 0 ... one of the solutions that i get is 4pi/3 + 2n*pi ... my book doesn't mention this... am i correct??

sin 4x - sin 2x = cos 3x ... one of the solutions that i get is pi/2 + 2n*pi/3 ... am i right?

sin x + sin 3x + sin 5x = 0 ... are pi/3 + 2n*pi and 2n*pi/3 also solutions to this eq. ??

LAST ONE :) ... sin θ + sin 3θ + sin 5θ + sin 7θ = 0 ... is 3*pi/4 + n*pi also a solution...

I really need to confirm the answers i am having my grand test tomorrow... i have tried them but not sure about the answers... i will be grateful forever if you confirm these... thanks :)

These are the same questions you asked in the other thread you posted, which makes this post of duplicate of the other. It is not good manners to post the same problem in multiple threads.

Did you read my response in your other thread? If you know how to find solutions of trig equations, you should be able to check whether a particular value of the variable makes your equation a true statement. If so, then that value is a solution of the equation. If not, then that value isn't a solution of the equation.