Use inverse trigonometric functions to find the solutions of

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. (Enter your answers as a comma-separated list.)

10 sin²x = sin x + 2;    [0, 2)

x=

Solution

10sin2x = -sinx+2 ---> If this is the question then it cannot be solved by alzebraically only graphical

solution is possible.

10sin^2x = -sinx+2

10sin^2x +sinx =2

10sin^2x +sinx -2 =0

solve by factorisation:

10sin^2x +5sinx -4sinx -2 =0

5sin(2sinx +1) -2(2sinx +1) =0

(5sinx-2)(2sinx+1) =0

sinx = 2/5

x =sin^-1(2/5) = 0.4115 radians

x = pi - sin^-1(2/5) = 2.7300 radians

sinx = -1/2

x= pi + sin^-1(1/2) = 3.6652

x = 2pi - sin^-1(1/2) = 5.7595

Solution : x= 0.4115, 2.7300 , 3.6652, 5.7595 radians