$\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------$

$(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s$

6 0

3 years ago

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Which equation has x=4 as the solution

dedylja [7]

Answer:

are you looking for example?

Step-by-step explanation:

If so,

X + 3 = 7

4 0

3 years ago