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Find all solutions of the given equation. csc^2 θ − 2 = 0
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It's an equation with a single trig function, so there's no harm in taking the 2 over. Now, do you know what csc is?
1/cos^2 x - 2 = 0 --> 1/cos^ x = 2 -> cos ^2 x = 1/2 cos x = V2/2 and cos x = -V2/2 a) cos x = V2/2 = cos Pi/4 --> x1 = ...... b) cos x = -V2/2 = cos 3Pi/4 -> x2 = ......
if your question is resolving (cos(x))^2 -2=0 then it doesn't have a solution because the cosinus is always between -1 and 1 so the square couldn't equal 0
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