sec 8.2 You are given an equation of the form sin x = k where k > 0.sin x = 0.1(a) Use the graph shown below to estimate (to the nearest 0.05) a root of the equation within the interval 0 sx < T/2.1.0 -0.90.80.70.60.50.4 -0.3-0.20.1 -0.10.20.3 0.4 0.50.60.70.80.91.01.11.21.31.4 1.5 1y = sin x, 0

Name: sec 8.2 You are given an equation of the form sin x = k where k > 0.si
Uploaded: 2024-03-20T06:46:55.000Z
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Description: sec 8.2 You are given an equation of the form sin x = k where k > 0.sin x = 0.1(a) Use the graph shown below to estimate (to the nearest 0.05) a root of the equation within the interval 0 sx < T/2.1.0 -0.90.80.70.60.50.4 -0.3-0.20.1 -0.10.20.3 0.4 0.

Step:-1 Part (a):- Given that sin(x) =0.1 Take sin(x) =y then y=0.1 Now, for y=0.1 we check correspond value on X-axis for y=0.1, then we get x=0.100 (using graph we can see) So, x=0.10 Part (b):- Using calculator we have sin-1(0.1)=0.1002 x=0.1002 Part (c):- Now, Since, we know sine is positive in Second quadrant. Note:- Since, we know Sinπ-x=sin(x) So, second angle is x=π-0.1002=3.0414x=3.0414 Part (d):- Now, given that sin(x)=-k that is, sin(x)=-0.1 Since, we know sine is negative in Third and fourth quadrant. So, In Third quadrant, x=π+0.1002=3.2418 In Fourth quadrant, x=2π-0.1002x=6.1830 So, x=3.2418, 6.1830 Answer:- Part (a):- x=0.10 Part (b):- x=0.1002 Part (c):-x=3.0414 Part (d):-x=3.2418, 6.1830