Question

How do you solve \[\sin 2x\sin x - \cos x = 0\]?

Answer 1

Hint: Here the question is related to the trigonometry, we use the trigonometry ratios and we are to solve this question. In this question we have to simplify the given trigonometric ratios to its simplest form. By using the trigonometry ratios and trigonometry formulas we simplify the given trigonometric function.

Complete step-by-step solution:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. In trigonometry the cosecant trigonometry ratio is the reciprocal to the sine trigonometry ratio. The secant trigonometry ratio is the reciprocal to the cosine trigonometry ratio and the cotangent trigonometry ratio is the reciprocal to the tangent trigonometry ratio.
The tangent trigonometry ratio is defined as \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] , The cosecant trigonometry ratio is defined as \[\csc x = \dfrac{1}{{\sin x}}\], The secant trigonometry ratio is defined as \[\sec x = \dfrac{1}{{\cos x}}\] and The tangent trigonometry ratio is defined as \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Now consider the given equation \[\sin 2x\sin x - \cos x = 0\]
By the formula of trigonometry, we have \[\sin 2x = 2\sin x\cos x\], substituting the formula to the given equation we have
\[ \Rightarrow 2\sin x\cos x(\sin x) - \cos x = 0\]
On simplifying we get
\[ \Rightarrow 2{\sin ^2}x\cos x - \cos x = 0\]
Take cos x as a common, the above inequality is written as
\[ \Rightarrow \cos x(2{\sin ^2}x - 1) = 0\]
Therefore we have
\[ \Rightarrow \cos x = 0\] and \[2{\sin ^2}x - 1 = 0\]
Hence
\[
   \Rightarrow x = {\cos ^{ - 1}}0 \\
   \Rightarrow x = \dfrac{\pi }{2} \\
 \]
Now consider \[2{\sin ^2}x - 1 = 0\]
This can be written as
 \[
   \Rightarrow 2{\sin ^2}x = 1 \\
   \Rightarrow {\sin ^2}x = \dfrac{1}{2} \\
   \Rightarrow \sin x = \pm \dfrac{1}{{\sqrt 2 }} \\
 \]
Hence we have
\[ \Rightarrow x = {\sin ^{ - 1}}\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right)\]
Therefore the value of x by the table of trigonometry ratios for the standard angles is given as
\[x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4}\]
These solutions belong to \[\left[ {0,\pi } \right]\].

Note: In the trigonometry we have six trigonometry ratios and 3 trigonometry standard identities. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, cosec or csc, sec and cot. The above question is also solved by using the standard trigonometry formulas on sine and cosine.