How to solve equations involving inverse trigonometric functions? Questions for grade 12 maths are presented along with detailed solutions. The solutions to the equations are also checked graphically. The left side and the right side of a given equation are graphed in the same system of coordinates and the approximation to the solution of the equation is given by the x coordinate of the point of intersection of the two graphs.

Question 1Solve for x the equation 3 arcsin(x) = π / 2.solutionDivide both sides of the equation by 3. arcsin(x) = (π / 2) / 3 arcsin(x) = π / 6 Apply sin to both sides and simplify. sin(arcsin(x)) = sin(π / 6) The above simplify to x = 1 / 2 Because of the domain of arcsin(x), we need to verify that the solution obtained is valid. x = 1 / 2 Right side of equation: 3 arcsin(1 / 2) = 3 (π6) = π / 2. Left side of equation: π / 2. The solution to the above equation is x = 1 / 2. The graphical approximation to the solution to the given equation is shown below. The x coordinate of the point of intersection of the graphs made up the left side and the right of the given equation is 0.5 which is the solution found analytically. .
Question 2Solve for x the equation 3 cot (arccos(x)) = 2.solutionDivide both sides of the given equation by 3 and simplify. cot (arccos(x)) = 2 / 3 Let A = arccos(x) and apply cos to both sides to obtain. cos (A) = cos(arccos(x)) = x Using definition of A above, the equation may be written as. cot (A) = 2 / 3 Use cot (A) = 2 / 3 to construct a right triangle and find cos(A). Find hypotenuse h first. . h = √(13) We now use the same triangle shown above to find cos (A). x = cos(A) = 2 / √(13) ≈ 0.55 The graphical approximation to the solution to the given equation is shown below. .
Question 3Solve for x the equation arcsin(x) = arccos(x).solutionApply sin function to both sides. sin(arcsin(x)) = sin(arccos(x)) Simplify left side using the identity sin(arcsin(A)) = A. x = sin(arccos(x)) let A = arccos(x) cos A = x sin(arccos(x)) = sin (A) = ~+mn~ √ (1  x^{ 2}) Use the above to rewrite the given equation in algebraic form. x = ~+mn~ √ (1  x^{ 2}) Square both sides. x^{ 2} = (1  x^{ 2}) 2 x^{ 2} = 1 x = ~+mn~ 1 / √(2) Because of the domain of the arccos function and we also squared both sides of the equation, we need to verify the solutions and eliminate any invalid (extraneous) ones. 1) x = 1 / √(2) left side: arcsin( 1 / √(2) ) = π / 4 right side: arccos( 1 / √(2) ) = π / 4 x = 1 / √(2) is a solution to the given equation. 2) x =  1 / √(2) left side: arcsin(  1 / √(2) ) =  π / 4 right side: arccos(  1 / √(2) ) = 3 &pi / 4 x =  1 / √(2) is not a solution to the given equation. The graphical approximation to the solution to the given equation is shown below. The x coordinate of the point of intersection 0.71 is close to 1/√2. .
Question 4Solve for x the equation arccos(x) = arcsin(x) + π / 2.solutionApply cos function to both sides. cos(arccos(x)) = cos( arcsin(x) + π / 2 ) Simplify left side using the identity cos(arccos(A)) = A. x = cos( arcsin(x) + π / 2 ) Expand the right side using the identity cos(a + b) = cos(a).cos(b)  sin(a)sin(b). x = cos( arcsin(x)) cos(π / 2)  sin( arcsin(x)) sin(π / 2) Use cos(π / 2) = 0 , sin( arcsin(x)) = x and sin(π / 2) = 1 to simplify the right side of the equation. x =  x 2 x = 0 x = 0 Verify the solution found. Left side: arccos(0) = π / 2 Right side: arcsin(0) + π / 2 = π / 2 x = 0 is a solution to the given equation The graphical approximation to the solution to the given equation is shown below. The x coordinate of the point of intersection is equal to 0 exactly as the value calculated analytically above. .
Question 5Solve for x the equation arccos(2x) = π/3 + arccos(x).solutionApply cosine function to both sides. cos(arccos(2x)) = cos(π/3+ arccos(x)) Simplify left side using the identity cos(arccos(A)) = A and expand the right side using the identity cos(a + b) = cos(a).cos(b)  sin(a)sin(b). left side: cos(arccos(2x)) = 2 x right side: cos(π/3+ arccos(x)) = cos(π/3)cos(arcos(x))  sin(π/3)sin(arccos(x)) = cos(π/3)x  sin(π/3)sin(arccos(x)) Rewrite sin(arccos(x)) and the right side of the equation as an algebraic expression. let A = arccos(x) , cos(A) = cos(cos(x)) = x sin(arcos(x)) = sin(A) = ~+mn~ √ (1  cos^{ 2}A) = ~+mn~ √ (1  x^{ 2}) right side: cos(π/3) x ~+mn~ sin(π/3)√ (1  x^{ 2}) Use cos(π/3) = 1 / 2 and sin(π/3) = √3 / 2 and rewrite the equation using algebraic expressions. 2 x = x / 2 ~+mn~ √3 / 2√ (1  x^{ 2}) Rewrite the equation with radical on the right side. 3 x / 2 = ~+mn~ (√3 / 2) √ (1  x^{ 2}) Square both sides of the equation and simplify. 9 x^{ 2} / 4 = [ ~+mn~ (√3 / 2)√ (1  x^{ 2}) ]^{ 2} 9 x^{ 2} / 4 = (3 / 4)(1  x^{ 2}) Solve for x. 12 x^{ 2} / 4 = 3 / 4 x^{ 2} = 1 / 4 x = ~+mn~ 1 / 2 Because of the domain of the arccos function and we also squared both sides of the equation, we need to verify the solutions and eliminate any invalid (extraneous) ones. 1) x = 1 / 2 left side: cos(arccos(2x)) = cos(arccos(2(1/2))) = cos(arccos(1)) = 0 right side: π/3 + arccos(x) = π/3+ arccos(1 / 2) = π/3 + π/3 = 2 π/3 x = 1 / 2 is not a solution to the given equation. 2) x =  1 / 2 left side: cos(arccos(2x)) = cos(arccos(2(  1/2))) = cos(arccos(  1)) = π right side: π/3 + arccos(x) = π/3+ arccos( 1 / 2) = π/3 + 2 π/3 = π x =  1 / 2 is a solution to the given equation. The graphical approximation to the solution to the given equation is shown below. The point of intersection has an x coordinate equal to 0.5 which is exactly the solution found analytically. . 