I am totally lost with this

solve:
2cos^2x=sinx+1

tan^2x=3tanx

verify identity:

1-cot^4x=2csc^2x-csc^4x

2cos^2(x) = sin(x) + 1
2(1 - sin^2(x)) = sin(x) + 1
2 - 2sin^2(x) = sin(x) + 1
0 = 2sin^2(x) + sin(x) - 1
0 = (2sin(x) - 1)(sin(x) + 1)
2sin(x) - 1 = 0 sin(x) + 1 = 0
2sin(x) = 1 sin(x) = -1
sin(x) = 1/2 x = 3pi/2
x = pi/6, 5pi/6

tan^2(x) = 3tan(x)
tan^2(x) - 3tan(x) = 0
tan(x)(tan(x) - 3) = 0
tan(x) = 0 tan(x) - 3 = 0
x = 0, pi tan(x) = 3
x = tan^-1(3)

1 - cot^4(x) = 2csc^2(x) - csc^4(x)
(1 - cot^2(x))(1 + cot^2(x)) = 2csc^2(x) - csc^4(x)
(1 - cot^2(x))(csc^2(x)) = 2csc^2(x) - csc^4(x)
([sin^2(x) / sin^2(x)] - [cos^2(x) / sin^2(x)])(csc^2(x)) = 2csc^2(x) - csc^4(x)
(sin^2(x) - cos^2(x))(csc^4(x)) = 2csc^2(x) - csc^4(x)
(sin^2(x) - [1 - sin^2(x)])(csc^4(x)) = 2csc^2(x) - csc^4(x)
(2sin^2(x) - 1)(csc^4(x)) = 2csc^2(x) - csc^4(x)
2csc^2(x) - csc^4(x) = 2csc^2(x) - csc^4(x)

I hope that all makes sense and that I did not use too many parentheses... there needs to be a more readable way to type equations.

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Quote:

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.there needs to be a more readable way to type equations.

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There is, it's called LaTex and this site implements it. Like so:




Click on quote to see the code I used to get that display.

prove that: cos^A=m^-1/n^-1,when sinA=msinA & tanA=ntanA