To simplify this expression, we can use the trigonometric identity cos²θ + sin²θ = 1.
Given expression: 2cos²(π/2+x) + √3 sinx = 0
Using the identity cos(π/2 + x) = -sin(x), we can rewrite the expression as follows:
2(-sinx)² + √3 sinx = 02sin²x + √3 sinx = 0
Now, let's factor out sinx:
sinx(2sinx + √3) = 0
Setting each factor equal to zero:
sinx = 0 or 2sinx + √3 = 0
If sinx = 0, then x = arcsin(0) = nπ, where n is an integer.
If 2sinx + √3 = 0:2sinx = -√3sinx = -√3/2
This occurs when x = -π/3 + 2nπ or x = 2π/3 + 2nπ, where n is an integer.
Therefore, the solutions are:x = nπ, where n is an integer, orx = -π/3 + 2nπ or x = 2π/3 + 2nπ, where n is an integer.
To simplify this expression, we can use the trigonometric identity cos²θ + sin²θ = 1.
Given expression: 2cos²(π/2+x) + √3 sinx = 0
Using the identity cos(π/2 + x) = -sin(x), we can rewrite the expression as follows:
2(-sinx)² + √3 sinx = 0
2sin²x + √3 sinx = 0
Now, let's factor out sinx:
sinx(2sinx + √3) = 0
Setting each factor equal to zero:
sinx = 0 or 2sinx + √3 = 0
If sinx = 0, then x = arcsin(0) = nπ, where n is an integer.
If 2sinx + √3 = 0:
2sinx = -√3
sinx = -√3/2
This occurs when x = -π/3 + 2nπ or x = 2π/3 + 2nπ, where n is an integer.
Therefore, the solutions are:
x = nπ, where n is an integer, or
x = -π/3 + 2nπ or x = 2π/3 + 2nπ, where n is an integer.