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Solve for x 3sinx - 2cosx^2x = 0
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3sinx -2cos²x + 2sin²x = 0 3sinx -2+4sin ²x=0 then factorise them
@sparkly16 Sorry to ask, but is this your question? \[3sinx - 2cos^2x = 0\]
3sinx−2cos² 2x=0
opps
3sinx−2cos 2x=0 tis one
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yes
Once again... Is it (i) or (ii)? \[(i)3sinx - 2cos^2x = 0\]\[(ii)3sinx - 2cos2x = 0\]
its (i)
oh so it is 3six x -2 +2 sin²x then sin x =0.5?
Use the identity \[sin^2x+cos^2x=1\] \[3sinx - 2cos^2x = 0\]\[3sinx - 2(1-sin^2x) = 0\]\[3sinx - 2 + 2sin^2x = 0\]\[2sin^2x + 3sinx - 2 = 0\]\[(2sinx-1)(sinx +2)=0\]\[2sin^2x + 3sinx - 2 = 0\]\[(2sinx-1)=0 \ or \ (sinx +2)=0\]\[sinx=\frac{1}{2} \ or \ sinx = -2 (rejected)\]So, sinx = 0.5 Take arc sin to find the value of x, can you?
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thank you so much! :)
Welcome :)
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