Question

Number of values of x which lie in [0, 2π] and satisfy the equation

`(cos  x/4 - 2sinx) sinx + (1 + sin  x/4 - 2cosx)cosx` = 0

Options

• 1

• 2

• 3

• 4

Answer 1

1

Explanation:

`(cos  x/4 - 2sinx) sinx + (1 + sin  x/4 - 2cosx)cosx` = 0

`\implies (sin x cos  x/4 + cos x sin  x/4) + cosx - 2(sin^2x + cos^2x)` = 0

`\implies sin(x + x/4) + cosx - 2(1)` = 0 `\implies sin  (5x)/4 + cosx` = 2

`\implies sin  (5x)/4` = cos x = 1

`\implies sin  (5x)/4` = 1 `\implies (5x)/4 = 2nπ + π/2` `\implies x = (8nπ)/5 + (2π)/5` and cos x = 1 `\implies` x = 2mπ

Thus we have `(8nπ)/5 + (2π)/5` = 2mπ `\implies m = (4n + 1)/5`

∴ n ∈ I, so m must be of the form m = 5k + 1

Hence the solution of the equation is x = 2(5k + 1) π, k ∈ I