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pre-calc trig find all solutions to the equation 7 sin^2x - 14 sin x + 2 = -5
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Replace \(\sin^2(x) \) with \(u^2\) and \(\sin(x)\) with \(u\). Quadratic equation right there
do I add the -5 to the 2?
bc quad equations always = 0 right?
Yes
then we end up with (u-1) (7u-7)
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then I switch the u back to sin x right?
mhm
no, first for \(u\)
wait, yeahso we end up with sinx=1 and sinx=1...?
Yes, :)
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And you can do it
ahh this is the part I get mixed up in :/
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