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# How many values of x in the interval [0,2π] satisfies the equation sin6x=1+cos43x ___

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## Maximum value of sin6 x is 1 and minimum value of 1+ cos4 3x is 1. ⇒ sin6 x ≤1 and sin6x = 1 + cos4 3x ≥ 1 ⇒ 1≤ sin6 x ≤ 1 ⇒ sin6 x = 1 ⇒ cos4 3x = sin6 x - 1 = 1 - 1 = 0 ⇒ sin6 x = 1 and cos4 3x = 0 ⇒x=(2n+1)π2 and 3x=(2m+1)π2. ⇒x=(2n+1)π2 and x=(2m+1)π6. We want to find the common values of (2n+1)π2and(2m+1)π6. For some value of m and n, both these expressions will be equal corresponding to common terms. ⇒(2n+1)π2=(2m+1)π6 ⇒2m+1=6n+3 ⇒2m=6n+2 ⇒ m = 3n + 1 This means for any value of n, there will be a corresponding m. ⇒ All the values of (2n + 1) π2 is also part of the sequence (2m + 1) π6. So the solution is x = (2n + 1) π2. The values in the interval [0, 2π] are π2and3π2 corresponding to n = 0 and n = 1. So the number of solutions in the interval [0, 2π] is 2.  Suggest Corrections  0      Similar questions  Related Videos   Solving Trigonometric Equations
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