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Question

Solve the following equations.
34cos2x+1=2cos2x+2.

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Solution

34cos2x+1=2cos2x+2
3=8cos22x+10cos2x+2
3=8(12sin2x)2+10(12sin2x)+2
[cos2x=12sin2x
squaring both sides
a=20+32sin4x52sin2x
take sin2x=t
then, t25232t+1132=0
t=52/32±(52/32)24×11/322
t=1.375
OR
0.25
sin2x=1.375
OR
0.25
sinx=±1.17 OR ±0.5
sinx=±1/2
x=π/6,2π/3,7π/6,5π/3

1124466_888290_ans_ec15a1b805e94933a718e3a0f91e11a7.jpg

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