Question

How do you solve $\left( {\sin x + 1} \right) - 2\cos x = 0$?

Answer 1

First move the cos part on the right side. After that square both sides and use the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ on the left side. After that use the identity ${\cos ^2}x = 1 - {\sin ^2}x$ and move all terms on one side to make a quadratic equation of $\sin x$. Then factorize the equation by factorization method to get the solution of $\sin x$. Finally, we will get the solution for the equation.

Complete step-by-step solution:
Given that,
$\Rightarrow \left( {\sin x + 1} \right) - 2\cos x = 0$
Move $2\cos x$ on the right side,
$\Rightarrow \left( {\sin x + 1} \right) = 2\cos x$
Now take square on both sides,
$\Rightarrow {\left( {\sin x + 1} \right)^2} = {\left( {2\cos x} \right)^2}$
Use the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ on left side,
$\Rightarrow {\sin ^2}x + 2\sin x + 1 = 4{\cos ^2}x$
Now, use the identity ${\cos ^2}x = 1 - {\sin ^2}x$,
$\Rightarrow {\sin ^2}x + 2\sin x + 1 = 4\left( {1 - {{\sin }^2}x} \right)$
Open the bracket and multiply the terms,
$\Rightarrow {\sin ^2}x + 2\sin x + 1 = 4 - 4{\sin ^2}x$
Move all terms on the left side and simplify,
$\Rightarrow 5{\sin ^2}x + 2\sin x - 3 = 0$
Now, we will compare the given equation with the standard quadratic equation which is given by $a{x^2} + bx + c = 0$.
On comparing we get the values $a = 5,b = 2,c = - 3$.
Now, we know that the quadratic formula is given as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values in the above formula we get
$\Rightarrow \sin x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 5 \times - 3} }}{{2 \times 5}}$
Now, on solving the obtained equation we get
$\Rightarrow \sin x = \dfrac{{ - 2 \pm \sqrt {4 + 60} }}{{10}}$
Add the terms in the square root,
$\Rightarrow \sin x = \dfrac{{ - 2 \pm \sqrt {64} }}{{10}}$
Simplify the terms,
$\Rightarrow \sin x = \dfrac{{ - 2 \pm 8}}{{10}}$
Now, we know that a quadratic equation has two roots. We can write the obtained equation as
$\Rightarrow \sin x = \dfrac{{ - 2 - 8}}{{10}}$ and $\sin x = \dfrac{{ - 2 + 8}}{{10}}$
Simplify the terms,
$\Rightarrow \sin x = \dfrac{{ - 10}}{{10}}$ and $\sin x = \dfrac{6}{{10}}$
Cancel out the common factors,
$\Rightarrow \sin x = - 1$ and $\sin x = \dfrac{3}{5}$
Take ${\sin ^{ - 1}}$ on both sides,
$\Rightarrow x = \dfrac{{3\pi }}{2}$ and $x = {\sin ^{ - 1}}\dfrac{3}{5}$
Hence, the solution of the equation is $\dfrac{{3\pi }}{2}$ and ${\sin ^{ - 1}}\dfrac{3}{5}$.

Note: In solving trigonometric equations we need to remember the formulas, the standard values of angles, and the identities because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identities as we can write cos in place of sin and sin in place of cos while solving for the general solution.