Question

Solve the following equation: \[\cos 2x + 3\sin x = 2\]

Answer 1

Hint: According to this particular question, we can use one of the basic formulas of trigonometry that is the double argument property of trigonometry and the formula for that is \[cox2x = 1 - 2{\sin ^2}x\].

Complete step-by-step solution:
The given equation is:
\[\cos 2x + 3\sin x = 2\]
First, we will substitute the above mentioned formula in the equation. We will substitute \[1 - 2{\sin ^2}x\]in place of \[cox2x\], and we get:
\[ \Rightarrow 1 - 2{\sin ^2}x + 3\sin x = 2\]
\[ \Rightarrow 1 - 2{\sin ^2}x + 3\sin x - 2 = 0\]
We will rearrange the terms, and we get:
\[ \Rightarrow - 2{\sin ^2}(X) + 3\sin (X) - 1 = 0\]
For a polynomial of the form \[a{x^2} + bx + c\], we will rewrite the middle term as a sum of those two terms whose product is \[a.c = - 2. - 1 = 2\]and whose sum is \[b = 3\].
Now, we will factor \[3\] out of \[3\sin (x)\], and we get:
\[ \Rightarrow - 2{\sin ^2}(x) + 3(\sin (x)) - 1 = 0\]
Now, we will rewrite \[3\] as \[1\] plus \[2\], and we get:
\[ \Rightarrow - {\sin ^2}(x) + (1 + 2)\sin (x) - 1 = 0\]
Now, we will apply the distributive property, and we get:
\[ \Rightarrow - 2{\sin ^2}(x) + 1\sin (x) + 2\sin (x) - 1 = 0\]
Now, we will multiply \[\sin (x)\] by \[1\]. We will group the first two terms and last two terms, and we get:
\[( - 2{\sin ^2}(x) + \sin (x) + 2\sin (x) - 1 = 0\]
Now, we will factor out the greatest common factor from each group, and we get:
\[ \Rightarrow \sin (x)( - 2\sin (x) + 1) - ( - 2\sin (x) + 1 = 0\]
Now, we will factor out the polynomial by factoring out the greatest common factor \[ - 2\sin x + 1\]
\[ \Rightarrow ( - 2\sin (x) + 1)(\sin (x) - 1) = 0\]
If any individual factor on the left side of the equation is equal to \[0\], then the entire expression will be equal to \[0\].
   \[\Rightarrow - 2\sin (x) + 1 = 0 \]
   \[\Rightarrow \sin (x) - 1 = 0 \]
We will set the first factor equal to \[0\], and we get:
\[ \Rightarrow - 2\sin (x) + 1 = 0\]
We will subtract \[1\] from both sides of the equation, and we get:
\[ \Rightarrow - 2\sin (x) = -1\]
We will divide each term by \[ - 2\] and simplify it. We will divide each term in \[ - 2\sin (x) = - 1\] by \[ - 2\], and we get:
\[ \Rightarrow\dfrac{{ - 2\sin (x)}}{{ - 2}} =\dfrac{{ - 1}}{{ - 2}}\]
We will cancel the common factor of \[ - 2\]:
\[ \Rightarrow \sin (x) =\dfrac{{-1}}{-2}\]
Dividing two negative values results in positive value
\[ \Rightarrow \sin (x) =\dfrac{1}{2}\]
We will take the inverse sine of both sides of the equation to extract \[x\] from inside the sine, and we get:
\[ \Rightarrow x = \arcsin (\dfrac{1}{2})\]
The exact value of \[\arcsin (\dfrac{1}{2})\]is \[\dfrac{\pi }{6}\]
\[ \Rightarrow x =\dfrac{\pi }{6}\]
The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from \[\pi \] to find solution in the second quadrant:
\[ \Rightarrow x = \pi -\dfrac{\pi }{6}\]
We will simplify \[\pi -\dfrac{\pi }{6}\] to write \[\dfrac{\pi }{1}\] as a fraction with a common denominator, multiply by \[\dfrac{6}{6}\]:
\[ \Rightarrow x =\dfrac{\pi }{1}.\dfrac{6}{6} -\dfrac{\pi }{6}\]
Now, multiply \[6\]by \[1\]:
\[ \Rightarrow x =\dfrac{{\pi .6}}{6} -\dfrac{\pi }{6}\]
We will combine the numerators over the common denominator:
\[ \Rightarrow\dfrac{{\pi .6 - \pi }}{6}\]
Simplify the numerator by moving \[6\] to left side of \[\pi \]:
\[ \Rightarrow x =\dfrac{{6.\pi - \pi }}{6}\]
Subtract \[\pi \]from \[6\pi \]:
\[ \Rightarrow x =\dfrac{{5\pi }}{6}\]
Now, we will find the period. The period of function can be calculated using \[\dfrac{{2\pi }}{{\left| b \right|}}\].
Replace \[b\] with \[1\] in the formula for period:
\[\dfrac{{2\pi }}{{\left| 1 \right|}}\]
The absolute value is the distance between a number and zero. The distance between \[0\] and \[1\] is
\[\dfrac{{2\pi }}{1}\]
Now, divide \[2\pi \] by \[1\]:
The period of the \[\sin x\] function is \[2\pi \]radians in both the directions:
\[ \Rightarrow x =\dfrac{\pi }{6} + 2\pi n,\dfrac{{5\pi }}{6} + 2\pi n\]
Set the next factor equal to \[0\]:
\[ \Rightarrow \sin (x) - 1 = 0\]
Add \[1\] to both sides of the equation:
\[ \Rightarrow \sin (x) = 1\]
Now, take the inverse sine of both the sides of the equation to extract \[x\]from inside the sine:
\[ \Rightarrow x = \arcsin (1)\]
The exact value of \[\arcsin (1)\] is \[\dfrac{\pi }{2}\]
\[ \Rightarrow x =\dfrac{\pi }{2}\]
The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from \[\pi \] to find the solution in the second quadrant.
\[ \Rightarrow x = \pi -\dfrac{\pi }{2}\]
Simplify \[\pi -\dfrac{\pi }{2}\]
To write \[\dfrac{\pi }{1}\] as a fraction with a common denominator, multiply \[\dfrac{2}{2}\]
\[ \Rightarrow x =\dfrac{\pi }{1}.\dfrac{2}{2} -\dfrac{\pi }{2}\]
Write each expression with a common denominator of \[2\], by multiplying each by an appropriate factor of \[1\]
Combine and multiply \[2\] by \[1\]:
\[ \Rightarrow x =\dfrac{{\pi .2}}{2} -\dfrac{\pi }{2}\]
Simplify the numerator:
\[ \Rightarrow x =\dfrac{\pi }{2}\]
The period of the sin(x) function is \[2\pi \] so values will repeat every \[2\pi \] radians in both directions.
\[ \Rightarrow x =\dfrac{\pi }{2} + 2\pi n\]for any integer \[n\]
The final solution is all the values that make \[( - 2\sin (x) + 1)(\sin (x) - 1) = 0\] is true
\[x =\dfrac{\pi }{6} + 2\pi n,\dfrac{{5\pi }}{6} + 2\pi n,\dfrac{\pi }{2} + 2\pi n\], for any integer \[n\].

Note: The above solution is easy and gets solved quickly. But there is one thing that is very important and we all should keep it in mind. The sign should be appropriately taken and substitution should be done accordingly.