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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
` -1.62gt=2` False. Interval `143.2lt=xlt=180` is not the solution.
For `323.2lt=xlt=360` , let x=330
`1.86gt=2` False. Interval `323.2lt=xlt=360` is not the solution.
For `270lt=xlt323.2` , let x = 280
Then, susbtitute the boundaries of the interval to check if it is included in the solution.
`2 gt= 2` True
Let x = 297.49
` 2.13gt=2` True
Answer: The interval `270lt=xlt=323.2`degrees satisties the equation of `sqrt5cos(x+63.4)gt=2` .
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