# Solve : 6 – 10cosx = 3$$sin^2x$$

## Solution :

we have, 6 – 10cosx = 3$$sin^2x$$

$$\therefore$$  6 – 10cosx = 3 – 3$$cos^2x$$

$$\implies$$  3$$cos^2x$$ – 10cosx + 3 = 0

$$\implies$$  (3cosx-1)(cosx-3) = 0  $$\implies$$  cosx = $$1\over 3$$ or cosx = 3

Since cosx = 3 is not possible as -1 $$\le$$ cosx $$\le$$ 1

$$\therefore$$  cosx = $$1\over 3$$ = cos($$cos^{-1}{1\over 3}$$)  $$\implies$$  x = 2n$$\pi$$ $$\pm$$ $$cos^{-1}{1\over 3}$$

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