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-sin^2x=2cos x-2
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\(\bf \textit{recall that}\\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\)
@jdoe0001 what comes next i am lost
well... aren't you just verifying identities?
no i am solving the equation
.... oh... shoot.. that identity there... is the wrong one =(
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its okay can you help me solve it though?
@primeralph
\(\bf \textit{recall that}\\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\qquad thus\\ \quad \\ -sin^2(x) = 2cos(x)-2\implies -(1-cos^2(x))=2cos(x)-2\\ \quad \\ -1+cos^2(x)=2cos(x)-2\implies cos^2(x)-2cos(x)+1=0 \) as you can see, it's just a quadratic
thank you i got it now
yw
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