5(1+cos2x1−cos2x−21+cos2x)=2cos2x+9
Let, cos2x=y
∴5(1+y1−y−21+y)=2y+9
⇒5[−y2−4y+1]=4y2+22y+18
⇒9y2+42y+13=0
⇒y=−31 or y=−313 (Not possible)
Now, cos4x=2cos22x−1
⇒cos4x=2(−31)2−1
⇒cos4x=−97
If 5(tan2x−cos2x)=2cos2x+9, then the value of cos4x is
Held on 2 Apr 2017 · Verified 6 Jul 2026.
−53
31
92
−97
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