If for x(0,2), _10 x+_10 x = -1 and _10( x+ x) = 12(_10n-1), n0, then the value of n is e…

Question

If for $x\in\left(0,\dfrac\pi2\right)$, $\log_{10}\sin x+\log_{10}\cos x = -1$ and $\log_{10}(\sin x+\cos x) = \dfrac12(\log_{10}n-1)$, $n>0$, then the value of $n$ is equal to:

Accepted Answer

$$\log_{10}(\sin x\cos x) = -1 \implies \sin x\cos x = \frac1{10}$$

$$2\log_{10}(\sin x+\cos x) = \log_{10}n - \log_{10}10 \implies (\sin x+\cos x)^2 = \frac n{10}$$

$$1+2\sin x\cos x = \frac n{10} \implies 1+\frac15 = \frac n{10} \implies n=12$$

Solutions · 1