Not bad, but I used the Pythagorean identity sec²x-tan²x=1 to solve the problem, because that is (secx+tanx)(secx-tanx)=1 so if the difference is 3 the sum would be 1/3 and from there you solve and get
4:50 I prefer using the formula: az² + bz + c = 0, z = sin x, a = 5, b = -1, c = -4. => sin x = z_1/2 = 1/10 +/- sqrt (1/100 + 4/5) = 0.1 +/- sqrt(0.81) = 0.1 +/- 0.9 = { -0.8, 1 }
This is a math Olympiad questions. function. Thankyou for responding. We will work on ur suggestion also after verifying ur answer. Pls subscribe our channel for more videos like this
@@classclips8 Yes, I've left out the trigonometry part, this was just the part with solving the quadratic polynomial equation using the formula z_1/2 = - b/2a +/- sqrt( (b/2a)² - c/a ), sometimes written as (- b +/- sqrt(D)) / 2a where D = b² - 4ac (discriminant). D < 0 => no real solution. D = 0 => one real solution. D > 0 => two real solutions. Always (for a != 0): two complex solutions.
You may have seen, that my solution was also written as "sin x = ... = { -0.8, 1 }". Like you did, I stroke out the solution sin x = 1 (cos x = 0) and didn't calculate arcsin(-4/5) = - 53.12° or - 0.9273 radiant = - 0.29517 π, because we were just looking for sin x = - 4/5.
Not bad, but I used the Pythagorean identity sec²x-tan²x=1 to solve the problem, because that is (secx+tanx)(secx-tanx)=1 so if the difference is 3 the sum would be 1/3 and from there you solve and get
Thanx for ur suggestion. We will verify ur method and use it in the future. Please subscribe our channel to get more mathematics
4:50 I prefer using the formula: az² + bz + c = 0, z = sin x, a = 5, b = -1, c = -4.
=> sin x = z_1/2 = 1/10 +/- sqrt (1/100 + 4/5) = 0.1 +/- sqrt(0.81) = 0.1 +/- 0.9 = { -0.8, 1 }
This is a math Olympiad questions. function. Thankyou for responding. We will work on ur suggestion also after verifying ur answer. Pls subscribe our channel for more videos like this
@@classclips8 Yes, I've left out the trigonometry part, this was just the part with solving the quadratic polynomial equation using the formula z_1/2 = - b/2a +/- sqrt( (b/2a)² - c/a ), sometimes written as (- b +/- sqrt(D)) / 2a where D = b² - 4ac (discriminant). D < 0 => no real solution. D = 0 => one real solution. D > 0 => two real solutions. Always (for a != 0): two complex solutions.
You may have seen, that my solution was also written as "sin x = ... = { -0.8, 1 }". Like you did, I stroke out the solution sin x = 1 (cos x = 0) and didn't calculate arcsin(-4/5) = - 53.12° or - 0.9273 radiant = - 0.29517 π, because we were just looking for sin x = - 4/5.