Solve (in radians) sec x = 2csc x
When I work it out, it gets to
2cot x - 1 = 0
cot x = 1/2
Would it be ok to use tan^-1 (2) ? I must demonstrate it with a graph as well... would the line go through Q I/III or II/IV ? Thank you!
sec x = 2csc x
1/cosx = 2/sinx
sinx = 2cosx
sinx/cosx = 2
tanx = 2
yes, I would use tan^-1 (2), but make sure your calculator is set to radians, (DRG key)
Also remember that the tangent is positive in quads I and III
tan^-1 (2) = appr 1.107
x = 1.107 or x = π+1.107= appr 4.249
To solve the equation sec x = 2csc x, we can start by converting sec x and csc x into their respective trigonometric functions.
Recall that sec x is equal to 1/cos x, and csc x is equal to 1/sin x. So, the equation becomes:
1/cos x = 2/(1/sin x)
Simplifying further, we can multiply both sides of the equation by cos x and sin x to get:
sin x = 2cos x
Now, let's divide both sides of the equation by cos x:
sin x / cos x = 2
Using the identity tan x = sin x / cos x, we have:
tan x = 2
To find the value of x, we can use the inverse tangent function (tan^-1) on both sides of the equation. So, x = tan^-1(2).
Now, to determine the quadrant(s) where the line intersects the graph, we can use the signs of sine and cosine.
In the first quadrant (Q1), both sine and cosine are positive. In the second quadrant (Q2), sine is positive, and cosine is negative. In the third quadrant (Q3), both sine and cosine are negative. In the fourth quadrant (Q4), sine is negative, and cosine is positive.
Since tan x = sin x / cos x, the value of tan^-1(2) is positive, therefore it lies in either Q1 or Q3.
To determine the exact quadrant, we can observe that the slope of the tangent function is positive for angles between 0 and pi/2. So, the line will intersect the graph in the first quadrant (Q1).
To demonstrate this on a graph, you can plot the tangent function and show that it intersects the line y = 2 at a positive angle between 0 and pi/2.
Remember to always verify the results using a calculator or software to find the exact value of x.