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PLEASE HELP.solve -sin^2 x = 2cos x-2
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\[\begin{array}{l} \text{Subtract }(2 \cos (x)-2)\text{ from both sides:} \\ -\sin ^2(x)-2 \cos (x)+2=0 \\ \text{Simplify trigonometric functions:} \\ \cos ^2(x)-2 \cos (x)+1=0 \\ \text{Substitute }u=\cos (x): \\ u^2-2 u+1=0 \\ \text{Factor the left hand side:} \\ (u-1)^2=0 \\ \text{Take the square root of both sides:} \\ u-1=0 \\ \text{Add }1\text{ to both sides:} \\ u=1 \\ \text{Substitute back for }u=\cos (x): \\ \cos (x)=1 \\ \text{Take the inverse cosine of both sides:} \\ x=0 \\\end{array}\]
and \[\text{x=360}\]
got some fascination over latex lately @.Sam. ?
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