 Hello and welcome to the session. In this session, we will solve trigonometric equations using inverse trigonometric functions with the help of technology or we can say with the help of model. Now, sometimes value of a trigonometric function is known and it is necessary to find the measure of an angle the concept of inverse functions can be applied to find the inverse of trigonometric functions. For example, we have sin x is equal to root 3 by 2, now here we have to find x. Now, by definition of inverse trigonometric functions, y is equal to sin of x if and only if x is equal to sin inverse of y. And minus pi by 2 is less than equal to x is less than equal to pi by 2. This means x varies from minus pi by 2 to pi by 2. Now, sin of x is equal to root 3 by 2 is equal to sin inverse of root 3 by 2 and this implies x is equal to the sin inverse of root 3 by 2 is equal to pi by 3. So, the required angle is pi by 3. Now, here we know at which angle sin is root 3 by 2 from the table of trigonometric functions. Now, here sin 60 degrees is root 3 by 2 that means sin pi by 3 is root 3 by 2. We can easily find the solution but this is not always the case. For example, if we have to solve the equation sin x is equal to 0.5738. We do not know the angle where 0.5738. Now, to solve this, we make use of scientific calculator. For example, on the scientific calculator we type if 0156 degrees now x is equal to 0.5738 implies 0.5738 we have got the value of 8 of 0.5738 that is we have got the value of sin inverse of 0.5738 with 015 approximately equal to 35 degrees. Now, let us discuss another example of the equation 3 cos 0 degree is less than equal to theta is less than equal to 180 degrees that means theta varies from 0 degree to 180 degrees. Now, 3 cos theta plus 2 is equal to 3 implies 3 cos theta is equal to 3 minus 2 that is equal to 1 and this gives cos theta is equal to 1 upon 3. And further this implies theta is equal to 1 upon 3 cos of 1 upon 3 and enter and we get this equal to 2 degrees equal to cos inverse 1 upon 3. Now, here from scientific calculator, we have got the value of 1 upon 3 which is equal to 17.52 degrees equal to 17.5 inverse 0 that is 0 degree is less than equal to theta is less than equal to 180 degrees. And the value of theta such that cos theta is equal to 1 upon 3. And now, let us discuss modeling object equations. Now, suppose a string and one experiment this position is above rest position and return to 6 inches below rest position once in every 6 into minus 6 into t. Equivalent models height above and below rest position in every 6 seconds find when the weighted ball first will be at a height of 3 inches and 4 inches above rest position. Now, it is a periodic function given by into t when we make use of inverse trigonometric functions. Let this be equation number 1, 3 in equation number 1 and we have 3 is equal to minus into t minus 6 is equal to cos of pi by 3 into t. Which further gives minus 1 upon 2 is equal to cos of pi by 3 into t. Now, by the definition of inverse trigonometric functions this implies inverse of minus 1 by 2 pi by 3 into t. Now, on first multiplication this implies t is equal to 3 upon pi into minus 1 by 2 this implies t is equal to 3 upon pi. Now, from scientific calculator we can find cos inverse of minus 1 by 2 and this is equal to 2 pi upon 3. This means cos inverse of minus 1 by 2 is equal to 2 pi upon 3 that is 120 degrees. Now, here 3 into 1 is 3 and pi will be cancelled with pi so t is equal to 2. This is equal to 3 we are getting t is equal to 2 it will reach at the height of 3 inches. And now we have well s is equal to now this is the equation 1 putting s is equal to 4 in equation 1 v hat into t by 3 into t and this gives cos 3 is equal to pi by 3 into t. Now, by definition of inverse trigonometric functions here we have cos inverse of minus 2 by 3 is equal to pi by 3 into t further. Now, here we will cos multiply and we have t is equal to cos inverse of minus 2 upon 3 by using scientific calculator. We have inverse of minus 2 by 3 is equal to 0 degrees which is equal to 132 degrees approximately. Now, we convert degrees into variance by multiplying 132 degrees by upon 180 degrees. Now, 12 into 11 is 132 and 12 into 15 is 180 so this is equal to 11 pi radians. So, 132 degrees is equal to 11 pi by 15 radians of minus 2 by 3 is approximately equal to by 15 radians. This implies t is approximately equal to 3 upon pi into 11 pi by 15 by whole. Now, 3 into 5 is 15 and pi will be conserved by t is approximately equal to 11 by 5 is approximately equal to 2.2 seconds. So, this means approximately it will reach the session we have learnt how to solve trigonometric equations using inverse trigonometric functions with the help of modelling. And this completes our session. Hope you all have enjoyed the session.