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No. of solutions of 16^(sin^2x)+16^(cos^...

No. of solutions of `16^(sin^2x)+16^(cos^2x)=10, 0 le x le 2 pi` is

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To find the number of solutions for the equation \( 16^{\sin^2 x} + 16^{\cos^2 x} = 10 \) in the interval \( 0 \leq x \leq 2\pi \), we can follow these steps: ### Step 1: Rewrite the equation We start with the equation: \[ 16^{\sin^2 x} + 16^{\cos^2 x} = 10 \] Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite \( 16^{\cos^2 x} \) as: \[ 16^{\cos^2 x} = 16^{1 - \sin^2 x} = \frac{16}{16^{\sin^2 x}} \] Thus, the equation becomes: \[ 16^{\sin^2 x} + \frac{16}{16^{\sin^2 x}} = 10 \] ### Step 2: Substitute \( t = 16^{\sin^2 x} \) Let \( t = 16^{\sin^2 x} \). Then, we can rewrite the equation as: \[ t + \frac{16}{t} = 10 \] ### Step 3: Multiply through by \( t \) To eliminate the fraction, multiply both sides by \( t \): \[ t^2 + 16 = 10t \] Rearranging gives us a quadratic equation: \[ t^2 - 10t + 16 = 0 \] ### Step 4: Solve the quadratic equation We can solve this quadratic equation using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = -10, c = 16 \): \[ t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2} \] This gives us: \[ t = \frac{16}{2} = 8 \quad \text{and} \quad t = \frac{4}{2} = 2 \] ### Step 5: Relate back to \( \sin^2 x \) Now we have two values for \( t \): 1. \( t = 8 \) implies \( 16^{\sin^2 x} = 8 \) \[ \sin^2 x = \frac{\log_2(8)}{4} = \frac{3}{4} \implies \sin x = \pm \frac{\sqrt{3}}{2} \] 2. \( t = 2 \) implies \( 16^{\sin^2 x} = 2 \) \[ \sin^2 x = \frac{\log_2(2)}{4} = \frac{1}{4} \implies \sin x = \pm \frac{1}{2} \] ### Step 6: Find solutions for \( \sin x = \pm \frac{\sqrt{3}}{2} \) The solutions for \( \sin x = \frac{\sqrt{3}}{2} \) are: \[ x = \frac{\pi}{3}, \quad \frac{2\pi}{3} \] The solutions for \( \sin x = -\frac{\sqrt{3}}{2} \) are: \[ x = \frac{4\pi}{3}, \quad \frac{5\pi}{3} \] ### Step 7: Find solutions for \( \sin x = \pm \frac{1}{2} \) The solutions for \( \sin x = \frac{1}{2} \) are: \[ x = \frac{\pi}{6}, \quad \frac{5\pi}{6} \] The solutions for \( \sin x = -\frac{1}{2} \) are: \[ x = \frac{7\pi}{6}, \quad \frac{11\pi}{6} \] ### Step 8: Count all solutions From the above, we have: - From \( \sin x = \pm \frac{\sqrt{3}}{2} \): 4 solutions - From \( \sin x = \pm \frac{1}{2} \): 4 solutions Thus, the total number of solutions in the interval \( 0 \leq x \leq 2\pi \) is: \[ 4 + 4 = 8 \] ### Final Answer The number of solutions is \( \boxed{8} \).
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