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Question

If 0x<2π , then the number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x=0, is

A
3
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B
5
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C
7
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D
9
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Solution

The correct option is C 7
Given equation iscosx+cos2x+cos3x+cos4x=0(cosx+cos3x)+(cos2x+cos4x)=02cos2xcosx+2cos3xcosx=02cosx(cos2x+cos3x)=02cosx(2cos5x2cosx2)=0cosxcos5x2cosx2=0cosx=0or cos5x2=0or cosx2=0Now, cosx=0x=π2, 3π2 [0x<2π]cos5x2=05x2=π2, 3π2, 5π2, 7π2, 9π2, 11π2,....x=π5, 3π5, π, 7π5, 9π5 [0x<2π]and cosx2=0x2=π2, 3π2, 5π2,......x=π [0x<2π]Hence, x=π2, 3π2, π, π5, 3π5, 7π5, 9π5

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