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3(sin2x+cos2x)+3(3−sin2x−cos2x)=28 show that either cos x = 0 or tan x = –1
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the question you have typed has an obvious mistake
it has no solution
expand the LHS and it all cancels to 9 so 9=28 a false statement , no solution
it's solvable. put 3-sin2x-cos2x=1-cos2x-sin2x=2. This on expansion gives us an equation in sin^2 x and cos^2 x, divide by cos^2 x and simplify to get the answer. It's tan x= -1.
k well I still dont understand what you did/are talking bout. but if you are correct, note that the answer the original asker gave said cosx =0 or tanx=-1 you divided out cos^2x , which is a variable , and u cnt divide by a variable
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