The general solution of sin10x+cos10x=2916cos42x is
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a
nπ4+π8,n∈I
b
nπ+π8,n∈I
c
nπ8+π4,n∈I
d
nπ+π4,n∈I
answer is A.
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Detailed Solution
Sin2x5+cos2x5=2916cos42x1−cos2x25+1+cos2x25=2916cos42xput cos2x=t1−t25+1+t25=29t416⇒24t4−10t2−1=0 By inspection, t2=12cos22x=121+cos4x2=12cos4x=04x=2n+1π2x=2n+1π8,n∈Z