# SOLUTION: Please help me solve these 2 questions: 14. If f(x)=3-2x and f^-1 denotes the inverse of f, then f^-1(o) is ? 15. What is the solution set of 2sin^2(x)-3sin(x)-2=0 where 0<=x<=2

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: Please help me solve these 2 questions: 14. If f(x)=3-2x and f^-1 denotes the inverse of f, then f^-1(o) is ? 15. What is the solution set of 2sin^2(x)-3sin(x)-2=0 where 0<=x<=2      Log On

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 Click here to see ALL problems on Trigonometry-basics Question 379535: Please help me solve these 2 questions: 14. If f(x)=3-2x and f^-1 denotes the inverse of f, then f^-1(o) is ? 15. What is the solution set of 2sin^2(x)-3sin(x)-2=0 where 0<=x<=2pi . Thank so much!Answer by J2R2R(94)   (Show Source): You can put this solution on YOUR website!14. f(x) = 3 - 2x which we will call y for ease of notation. So y = f(x) y = 3 - 2x; 2x = 3 - y; x = (3 - y)/2. Now x = f^-1(y) = (3 - y)/2 So f^-1(x) = (3 - x)/2 Therefore f^-1(0) = (3 - 0)/2 = 1.5 and we can see f(1.5) = 3 - 3 = 0 which returns the original value given. 15. 2sin^2(x) - 3sin(x) - 2 = 0 This is a quadratic equation which can be factorised as follows: (2sin(x) + 1)(sin(x) - 2) = 0 We have two possible solutions: 2sin(x) + 1 = 0 and sin(x) - 2 = 0; giving 2sin(x) = -1 and sin(x) = 2. Sin(x) = 2 has no solution, so we are left with sin(x) = -0.5 which we have to solve for 0 <= x <= 2pi You should be able to see that for sines of 0.5 we have 30 and 150 degrees but because they are negative they are in the 180 to 360 degrees zone. 30 degrees is pi/6, so we have solutions pi + pi/6 and 2pi - pi/6 which are 7pi/6 and 11pi/6 Hope this helps.