# find the range of x between 0and 360 degree that satisfy the inequality √ 5 cos (x+63.4 degree)>= 2.

lemjay | High School Teacher | (Level 2) Senior Educator

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`sqrt5 cos(x+63.4) gt= 2`

To solve for x, change the inequality sign to to equal sign.

`sqrt5 cos(x+63.4) = 2`

`cos (x + 63.4) = 2/sqrt5`

Rationalize right side.

`cos(x+63.4) = (2sqrt5)/5`

`x+63.4 = cos^(-1) (2sqrt5)/5`

`x+63.4 =26.6 `

`x = -36.8`

Since the value of angle x is negative,then, from the positive x-axis rotate `36.8^o ` clockwise. Also, from negative x-axis, rotate `36.8^o` clcokwise. So, angle x is located below the (+)x-axis and above (-)x-axis.

In a unit circle chart, angle x is located at the second and fourth quadrant. Its equivalent positive angle is:

`x= 180 - 36.8 = 143.2^o`  and `x=360-36.8 = 323.2^o`

These two values of x serves as a boundary in the quadrant where the angle belongs. This indicates that in each quadrant, there are two intervals.

Assign values of x to each intervals and substitute it to the original inequality equation to determine if the resulting condition is true.

For `90lt=xlt=143.2` , let x=100

`sqrt5 cos (100+63.4)gt=2`

`-2.14gt= 2`  False. Interval `90lt=xlt=143.2` is not the solution.

For `143.2lt=x<=180` , let x = 160

`sqrt5cos(160+63.4)gt=2`

` -1.62gt=2`   False. Interval `143.2lt=xlt=180` is not the solution.

For `323.2lt=xlt=360` , let x=330

`sqrt5cos(330+63.4)gt=2`

`1.86gt=2`  False. Interval `323.2lt=xlt=360` is not the solution.

For `270lt=xlt323.2` , let x = 280

`sqrt5cos(280+63.4)gt=2`

`2.14gt=2`  True.

Then, susbtitute the boundaries of the interval to check if it is included in the solution.

Let x=270

`sqrt5cos(270+63.4)gt=2`

`2 gt= 2` True

Let x = 297.49

`sqrt5cos(270+63.4)gt=2`

` 2.13gt=2`  True

Answer: The interval `270lt=xlt=323.2`degrees satisties the equation of `sqrt5cos(x+63.4)gt=2` .