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Question: Solve: \[\sin 2x = \sqrt 2 \cos x\] A. \[x = n\pi + \dfrac{\pi }{2}\] B. \[x = 2n\pi + \dfrac{\p...

Solve: sin2x=2cosx\sin 2x = \sqrt 2 \cos x
A. x=nπ+π2x = n\pi + \dfrac{\pi }{2}
B. x=2nπ+π4x = 2n\pi + \dfrac{\pi }{4}
C. x=2nππ4x = 2n\pi - \dfrac{\pi }{4}
D. x=nππ2(nI)x = n\pi - \dfrac{\pi }{2}(n \in I)

Explanation

Solution

Trigonometry is one of those divisions in mathematics that helps in finding the angles and missing sides of a triangle with the help of trigonometric ratios. We are given a trigonometry equation, sin2x=2cosx\sin 2x = \sqrt 2 \cos x . First, we will use the transposition method and make this equation equal to zero. Then, we will use the trigonometry identities, sin2x=2sinxcosx\sin 2x = 2\sin x\cos x . And then, we will also use some values too of cos and sin to find the value of x. Thus, we will get the final output.

Complete step by step answer:
Given that, sin2x=2cosx\sin 2x = \sqrt 2 \cos x
By using transposition method, we will move the RHS term to LHS and we will get,
sin2x2cosx=0\Rightarrow \sin 2x - \sqrt 2 \cos x = 0
We know that, sin2x=2sinxcosx\sin 2x = 2\sin x\cos x
Substituting this value in the above equation, we will get,
2sinxcosx2cosx=0\Rightarrow 2\sin x\cos x - \sqrt 2 \cos x = 0
Taking cosx common and evaluating this equation, we get,
cosx(2sinx2)=0\Rightarrow \cos x(2\sin x - \sqrt 2 ) = 0
We will equate this both with zero.Angles can be in Degrees or Radians. Unit circle is a circle with a radius of 1 with its centre at 0.With the use of this unit circle, we will find the values of x.

First, cosx=0\cos x = 0
x=π2\Rightarrow x = \dfrac{\pi }{2} or x=3π2 \Rightarrow x = \dfrac{{3\pi }}{2}
Next, 2sinx2=02\sin x - \sqrt 2 = 0
2sinx=2\Rightarrow 2\sin x = \sqrt 2
sinx=22\Rightarrow \sin x = \dfrac{{\sqrt 2 }}{2}
sinx=12\Rightarrow \sin x = \dfrac{1}{{\sqrt 2 }}
x=π4\Rightarrow x = \dfrac{\pi }{4} or x=3π4x = \dfrac{{3\pi }}{4}
Thus, x=nπ±π2x = n\pi \pm \dfrac{\pi }{2}, x=2nπ+π4x = 2n\pi + \dfrac{\pi }{4}
Hence, for the given trigonometric equation sin2x=2cosx\sin 2x = \sqrt 2 \cos x, the value of x is x=2nπ+π4x = 2n\pi + \dfrac{\pi }{4}.

Hence, the correct answer is option B.

Note: In trigonometry, we will study the relationship between the sides and angles of a right-angled triangle. The basics of trigonometry define three primary functions which are sine, cosine and tangent. Trigonometry can be divided into two sub-branches called plane trigonometry and spherical geometry. The trigonometric ratios of a triangle are also called the trigonometric functions. Its applications are in various fields like oceanography, seismology, meteorology, physical sciences, astronomy, acoustics, navigation, electronics, etc.