solve for x for angles between o and 360 degrees
(i)cos2x+sinx=0
cos2x+sinx=0
(1 - 2sin^2 x) + sinx = 0
2sin^2 x - sinx - 1 = 0
(2sinx +1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
for sinx = -1/2, x is in quads III or IV
x = 210° or 330°
for sinx = 1
x = 90°
x = 90° , 210° or 330°
To solve the equation cos(2x) + sin(x) = 0 for angles between 0 and 360 degrees, you can use algebraic manipulation and trigonometric identities.
Let's start by rearranging the equation:
cos(2x) + sin(x) = 0
Now, replace cos(2x) with its equivalent form using the double angle formula:
1 - 2sin^2(x) + sin(x) = 0
Rearrange the terms and set the equation equal to zero:
-2sin^2(x) + sin(x) + 1 = 0
To solve this quadratic equation, let's substitute sin(x) with a new variable, let's say u. The equation becomes:
-2u^2 + u + 1 = 0
Now, you can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, the equation does not factor easily, so we will use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
For the equation -2u^2 + u + 1 = 0, the coefficients are:
a = -2, b = 1, c = 1
Plugging these values into the quadratic formula, we get:
u = (-1 ± √(1 - 4*(-2)*1)) / (2*(-2))
u = (-1 ± √(1 + 8)) / (-4)
u = (-1 ± √9) / (-4)
u = (-1 ± 3) / (-4)
This gives us two possible solutions for u:
1) u = ( -1 + 3 ) / (-4) = 2 / (-4) = -1/2
2) u = ( -1 - 3 ) / (-4) = -4 / (-4) = 1
Now, remember that we initially substituted u with sin(x), so we can rewrite these solutions as:
1) sin(x) = -1/2
2) sin(x) = 1
To find the values of x, we can use the inverse sine function (sin^-1) or reference angles.
1) sin(x) = -1/2:
The angles in which sine is equal to -1/2 are -30 degrees, which is the reference angle, and its supplementary angle 180 - (-30) = 210 degrees. However, we need to consider the quadrants where sin(x) is negative (i.e., Quadrant III and IV).
In Quadrant III, x = 180 + (-30) = 150 degrees.
In Quadrant IV, x = 360 - (-30) = 390 degrees. Since this is outside the defined range between 0 and 360 degrees, we discard this solution.
Therefore, x = 150 degrees.
2) sin(x) = 1:
There is only one angle in the range of 0 to 360 degrees where sin(x) is equal to 1, and that is x = 90 degrees.
So, the solutions for the equation cos(2x) + sin(x) = 0 in the range between 0 and 360 degrees are x = 150 degrees and x = 90 degrees.