If 2+=1, 2, then 7+6 is equal to:

Question

If $2\cos	heta+\sin	heta=1$, $	heta
e\dfrac\pi2$, then $7\cos	heta+6\sin	heta$ is equal to:

Accepted Answer

Squaring: $4\cos^2\theta+\sin^2\theta+4\sin\theta\cos\theta=1 \implies 3\cos^2\theta+4\sin\theta\cos\theta=0 \implies \cos\theta(3\cos\theta+4\sin\theta)=0$

Since $\theta\ne\tfrac\pi2$, $\cos\theta\ne0$, so $3\cos\theta+4\sin\theta=0 \implies \tan\theta=-\tfrac34$.

Using $\sec^2\theta=1+\tan^2\theta=\tfrac{25}{16}$, $\cos\theta=\pm\tfrac45$; testing against the original equation, $\cos\theta=\tfrac45,\sin\theta=\tfrac35$ satisfies it (the other sign does not).

$$7\cos\theta+6\sin\theta = 7\left(\frac45\right)+6\left(\frac35\right) = \frac{28}5+\frac{18}5 = \frac{46}5$$

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