Question

find the √3sinx+cosx=2​

Answer 1

Trigonometric Equation Solving

To solve: $\sqrt{3}\:sinx+cosx=2$

Step-by-step explanation:

Now, $\sqrt{3}\:sinx+cosx=2$

Divide both sides by $2$.

$\Rightarrow \frac{\sqrt{3}}{2}\:sinx+\frac{1}{2}\:cosx=1$

Since $cos30^{\circ}=\frac{\sqrt{3}}{2},\:sin30^{\circ}=\frac{1}{2}$, we get

$\Rightarrow cos30^{\circ}\:sinx+sin30^{\circ}\:cosx=1$

Use the formula: $sinA\:cosB+cosA\:sinB=sin(A+B)$

$\Rightarrow sin(x+30^{\circ})=1$

The value of $sin90^{\circ}$ is $1$.

$\Rightarrow sin(x+30^{\circ})=sin90^{\circ}$

Take $arcsine$ or $sin^{-1}()$ on both sides.

$\Rightarrow x+30^{\circ}=90^{\circ}$

$\Rightarrow x=60^{\circ}$

Answer:

Therefore the value of x is 60°.