Question

11. Solve: sin x + 3 cos x = V2 ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{sin(x)+3\,cos(x)=\sqrt{2}}$

$\sf{\implies\,sin(x)=\sqrt{2}-3\,cos(x)}$

$\sf{\implies\,sin^{2}(x)=(\sqrt{2}-3\,cos(x))^{2}}$

$\sf{\implies\,sin^{2}(x)=2+9\,cos^{2}(x)-6\sqrt{2}\,cos(x)}$

$\sf{\implies\,1-cos^{2}(x)=2+9\,cos^{2}(x)-6\sqrt{2}\,cos(x)}$

$\sf{\implies\,1=2+10\,cos^{2}(x)-6\sqrt{2}\,cos(x)}$

$\sf{\implies\,10\,cos^{2}(x)-6\sqrt{2}\,cos(x)+1=0}$

$\sf{\implies\,cos(x)=\dfrac{6\sqrt{2}\,\pm\,\sqrt{(-6\sqrt{2})^2-4\cdot10\cdot1}}{2\cdot10}}$

$\sf{\implies\,cos(x)=\dfrac{6\sqrt{2}\,\pm\,\sqrt{72-40}}{20}}$

$\sf{\implies\,cos(x)=\dfrac{6\sqrt{2}\,\pm\,\sqrt{32}}{20}}$

$\sf{\implies\,cos(x)=\dfrac{6\sqrt{2}\,\pm\,4\sqrt{2}}{20}}$

$\sf{\implies\,cos(x)=\dfrac{6\sqrt{2}\,+\,4\sqrt{2}}{20}\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,cos(x)=\dfrac{6\sqrt{2}\,-\,4\sqrt{2}}{20}}$

$\sf{\implies\,cos(x)=\dfrac{10\sqrt{2}}{20}\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,cos(x)=\dfrac{2\sqrt{2}}{20}}$

$\sf{\implies\,cos(x)=\dfrac{\sqrt{2}}{2}\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,cos(x)=\dfrac{\sqrt{2}}{10}}$

$\sf{\implies\,cos(x)=\dfrac{1}{\sqrt{2}}\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,cos(x)=\dfrac{\sqrt{2}}{10}}$

$\sf{\bullet\bold{\green{\,\,Solving\,\,\,the\,\,\,first\,\,\,condition:}}}$

We have,

$\sf{\implies\,\blue{cos(x)=\dfrac{1}{\sqrt{2}}}}$

$\sf{\implies\,cos(x)=cos\bigg(\dfrac{\pi}{4}\bigg)}$

$\sf{\implies\,x=2n\pi\,\pm\,\dfrac{\pi}{4}\,\,\,\,\,\,\,\,\,\forall\,n\in\mathbb{Z}}$

$\sf{\bullet\bold{\green{\,\,Solving\,\,\,the\,\,\,second\,\,\,condition:}}}$

We have,

$\sf{\implies\,\blue{cos(x)=\dfrac{\sqrt{2}}{10}}}$

$\sf{\implies\,cos(x)=cos(\alpha)}$

$\sf{where,\,\,\alpha=tan^{-1}(7)}$

So,

$\sf{\implies\,x=2m\pi\,\pm\,\alpha\,\,\,\,\,\,\,\,\,\forall\,m\in\mathbb{Z}}$