Is 180 a Valid Solution for the Trigonometry Equation in Part B) ii?

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The discussion focuses on verifying whether 180 is a valid solution for the equation sin2x = 2 - 2cos2x. The user initially derived that tanx = 1/2 but realized that 180 is not a solution due to dividing by sin2x, which can equal zero. It was clarified that when dividing by a trigonometric function, one must ensure it is not zero to avoid losing potential solutions. The user confirmed that x = 180 is indeed a solution by substituting it back into the original equation, showing that both sides equal zero. The conversation emphasizes the importance of checking for zero values when manipulating trigonometric equations.
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in part b) i of a question, i was asked to prove that (1-cos2x)/sin2x is equivilent to tanx and so i did.

then in part b) ii i was asked to varify that 180 is a solution of x for:
sin2x = 2 - 2cos2x

i took the 2 out; sin2x = 2(1 - cos2x) and saw similarities to part i,
so i divided by 2sin2x to get
(1-cos2x)/sin2x = 1/2
which using part i i deduced that tanx = 1/2

but 180 isn't the solution of this. (26.6 is)

where did I go wrong, was it because I divided through by a the 2sin2x when i shouldn't have?
in my https://www.physicsforums.com/showthread.php?t=165036" it was apparently ok to divide by a trig funtion, if this is the problem here why does it apply to this and not my other question

NB part a of the question seems irrelevant.
part b) iii however asks me to find the other 2 solutions, of which 26.6 is one of the answers
so how do i get to 180 and the other solution for part b) iii
 
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It's ok to divide by a trig function unless in the end that trig function turns out to be zero. If you put 180 in sin(2x) you get zero. So the original equation has zeros that the divided equation does not. Eg. x*(x-1)=0 has the solutions 0,1. (x-1)=0 has only 1.
 
ok so as sin2x = 0 for x=180, when you divide it goes.

so if i can't do that then how do i varify that x=180

i've already found out how to get the other solution, i forgot to mention that the interval is (0, 360)
 
When you divide by something make a note to yourself to check in the end that the thing you divided by is not zero, and if it is zero, then check that it is not an extra root. So you need to solve both tan(x)=1/2 and sin(2x)=0 and then check if all such values are really roots. Ie substitute them into sin2x = 2 - 2cos2x.
 
To verify x=180 is a solution
LHS is sin(360)=0
RHS is 2 -2cos(360)=2-2=0
LHS=RHS so x=180 is a solution

to obtain x=180 as a solution

(1-cos2x)=(tanx)(sin2x)

sin2x=2(1-cos2x)
sin2x=2(tanx)(sin2x)
sin2x-2tan(x)(sin2x)=0
sin2x(1-2tanx)=0

sin2x=0 or 1-2tanx=0
 
ahh its all clear now, would either one of the above proofs for x=180 be valid as an answer or will the latter have to be shown?
 
Depends on the wording of the question

verify that x=180 is a solution of ... use the first method

show that x=180 is a solution of... use the second method.
 
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