Question

How do you solve $2-2{{\cos }^{2}}x=5\sin x+3?$

Answer 1

Hint: We can solve this by using different trigonometric identities. There are various methods of solving. Here we can use the use Pythagorean identities i.e. ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
${{\sin }^{2}}x=1-{{\cos }^{2}}x$ by using this identity we can simplify the equation then shift the right side of the equation to the left side and carefully change the sign.

Complete step-by-step answer:
We have to solve this equation
$\Rightarrow$ $2-2\left( {{\cos }^{2}}x \right)=5\sin x+3$
We can write $2-2{{\cos }^{2}}x$ as $2\left( 1-{{\cos }^{2}}x \right)$
$\Rightarrow$ $2\left( 1-{{\cos }^{2}}x \right)=5\sin x+3$
Shift right side equation in the left side.
$\Rightarrow$ $2\left( 1-{{\cos }^{2}}x \right)=5\sin x+3$
We can write $1-{{\cos }^{2}}x={{\sin }^{2}}x$ because we know the identity ${{\sin }^{2}}x=\left( 1-{{\cos }^{2}}x \right)$
It is known as Pythagoras identity.
$\Rightarrow$ $2{{\sin }^{2}}x-5\sin x-3=0$
$\Rightarrow$ $2{{\sin }^{2}}x-6\sin x+\sin x-3$
We will factorise this equation so we gat,
$\Rightarrow$ $2\sin x\left( \sin x-3 \right)+\left( \sin x-3 \right)=0$
$\Rightarrow$ $\left( 2\sin x+1 \right)\left( \sin x-3 \right)=0$
Therefore $2\sin x=1$ and $\sin x=3$
As we know $\sin x$ lies between $\left[ -1,1 \right]$
So, $\sin x\ne 3$
Therefore $2\sin x=-1$ or $x=-\dfrac{1}{2}$
We know from the trigonometry formula.
$\Rightarrow$ $\sin x=-\dfrac{1}{2}$
$\Rightarrow$ $\sin \left( -30 \right)=-\dfrac{1}{2}$

And $\sin 210=-\dfrac{1}{2}$ we applied trigonometry table so the solution is $x=210{}^\circ $ and $x=-30{}^\circ $

Additional Information:
We can solve this by another method.
The method is given by
$\Rightarrow$ $2-2{{\cos }^{2}}x=5\sin x+3$
We can write $2-2{{\cos }^{2}}x$ as $2\left( 1-{{\cos }^{2}}x \right)$
$\Rightarrow$ $2\left( 1-{{\cos }^{2}}x \right)=5\sin x+3$
As using Pythagoras identity we can write
$\Rightarrow$ $1-{{\cos }^{2}}x={{\sin }^{2}}x$
Now, $2{{\sin }^{2}}x=5x+3=0$
Shift the right side equation to the left.
$\Rightarrow$ $2{{\sin }^{2}}x-5\sin x-3=0$
Now solve this quadratic equation for $\sin x$ by the improved quadratic formula in graphic form.
$\Rightarrow$ $D={{d}^{2}}={{b}^{2}}-4ac$
We know the quadratic equation form
$\Rightarrow$ $ax+by+c$
According to this we have equation
$\Rightarrow$ $2{{\sin }^{2}}x-5\sin x=3$
Where, $a=2$
$b=-5$
$c=-3$
Put the values in formula
$\Rightarrow$ ${{d}^{2}}={{b}^{2}}-4ac={{\left( -5 \right)}^{2}}-4\times 2\times \left( -3 \right)$
$\Rightarrow$ ${{d}^{2}}=25+24=29$
So, the value of $d=\pm 7$
There are two real roots $\sin x=\dfrac{-b}{2a}\pm \dfrac{d}{2a}$
(a) $\sin x=\dfrac{-b}{2a}\pm \dfrac{d}{2a}=\dfrac{-5}{2\times 2}\pm \dfrac{7}{2\times 2}=\dfrac{5}{4}\pm \dfrac{7}{4}$
$\Rightarrow$ $\sin x=\dfrac{5\pm 7}{4}=\dfrac{12}{4}=3$
$\Rightarrow$ $\sin x=3$ as rejected as $>1$ and
(b) $\sin x=\dfrac{-2}{4}=-\dfrac{1}{2}$
Trigonometry table of special arcs and unit circle.
$\Rightarrow$ $\sin x=-\dfrac{1}{2}>x=-\dfrac{\pi }{6}$ and $x=-\dfrac{5\pi }{6}$
There co terminals are $\dfrac{11\pi }{6}$ and $\dfrac{7\pi }{6}$
General answers.
$\Rightarrow$ $x=\dfrac{7\pi }{6}+2k\pi $
$\Rightarrow$ $x=\dfrac{11\pi }{6}+2k\pi $

Note:
The integer $k$ could be a positive or negative whole number of $0.$ If $k$ is negative we are subtracting from the basic in method $2.$ Memorise the trigonometric formula. Because this will lead you to the perfect solution. I would highly recommend memorization.