Question

How do you solve $1 + \sin x = 2{\cos ^2}x$ and find all exact general solutions?

Answer 1

Hint: Here the given expression is in the form of sine and cosine terms. So here we will convert the whole equation in one trigonometric identity that is convert the given equation using the correlation between sine and cosine and then find the resultant required value for “x”.

Complete step-by-step solution:
Take the given expression: $1 + \sin x = 2{\cos ^2}x$
Use the identity relating between the sine and cosine. ${\sin ^2}x + {\cos ^2}x = 1$
Frame the equation to substitute the value for cosine, ${\cos ^2}x = 1 - {\sin ^2}x$
Place the above value in the given expression.
$\Rightarrow 1 + \sin x = 2(1 - {\sin ^2}x)$
Multiply the term outside the bracket with the terms inside the bracket.
$\Rightarrow 1 + \sin x = 2 - 2{\sin ^2}x$
Move all the terms from the right hand side of the equation to the left hand side of the equation. When you move any term from one side to another then the sign of the term also changes. Positive term become negative and the negative term becomes positive.
$\Rightarrow 1 + \sin x - 2 + 2{\sin ^2}x = 0$
Take like terms together.
$\Rightarrow 2{\sin ^2}x + \sin x - \underline {2 + 1} = 0$
When you simplify between the two like terms and with the two opposite signs then you have to do subtraction and then give a sign of bigger number to the resultant value.
$\Rightarrow 2{\sin ^2}x + \sin x - 1 = 0$
Split the middle term in the above equation.
$\Rightarrow 2{\sin ^2}x + \underline {2\sin x - \sin x} - 1 = 0$
Make the pair of first two terms and the last two terms.
$\Rightarrow \underline {2{{\sin }^2}x + 2\sin x} - \underline {\sin x - 1} = 0$
Find the common factors from the paired terms.
$\Rightarrow 2\sin x(\sin x + 1) - 1(\sin x + 1) = 0$
Take the common factor, common from the above equation.
$\Rightarrow (\sin x + 1)(2\sin x - 1) = 0$
We get two equations,
$\Rightarrow \sin x + 1 = 0$and $2\sin x - 1 = 0$
Make the term with the variable “x” the subject-
$\Rightarrow \sin x = \dfrac{1}{2}$and $\sin x = - 1$
Using the All STC rule,
$\Rightarrow x = \dfrac{\pi }{6} + 2\pi n,\dfrac{{5\pi }}{6} + 2\pi n,\dfrac{{3\pi }}{2} + 2\pi n$
Where “n” is any integer

Note: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ).