2 into sin to the power 6 theta + cos to the power 6 theta minus 3 into 10 to the power 4 theta + cos to the power 4 theta + 1 is equal to zero
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22
Answer:
2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1
=2[(sin²θ)³+(cos²θ)³]-3[(sin²θ)²+(cos²θ)²]+1
=2[(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)]-3[(sin²θ+cos²θ)²-2sin²θcos²θ]+1
=2[(1)³-3sin²θcos²θ(1)]-3[(1)²-2sin²θcos²θ]+1 [∵, sin²θ+cos²θ=1]
=2-6sin²θcos²θ-3+6sin²θcos²θ+1
=2-3+1
=3-3
=0
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Answered by
1
Step-by-step explanation:
Answer is -1
on solving the equations by breaking it into hole cube and hole square we get the answer as -1
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