Solve: Sin 2x + sin 4x + sin 6x = 0. - Vidyadip

Question

Solve: Sin 2x + sin 4x + sin 6x = 0.

✓

Answer

We have,
$sin 2 x+sin 4 x+sin 6x = 0$
$\Rightarrow$ $sin 4 x+(sin 2 x+sin 6x) = 0$
$\Rightarrow$ $sin 4 x+2 sin 4 x\ cos 2x = 0$
$\Rightarrow$ $sin 4 x(1+2 cos 2x) = 0$
$\Rightarrow$ sin 4 x=0 or ,1+2 cos 2 x=0 $\Rightarrow$ sin 4x = 0 or $\cos 2 x=-\frac{1}{2}$
Now, $\sin 4 x=0 \Rightarrow 4 x=n \pi, n \in Z \Rightarrow x=\frac{n \pi}{4}, n \in Z$
And, $\cos 2 x=-\frac{1}{2}$
$\Rightarrow \quad \cos 2 x=\cos \frac{2 \pi}{3}$
$\Rightarrow \quad 2 x=2 m \pi \pm \frac{2 \pi}{3}, m \in Z$
$\Rightarrow \quad x=m \pi \pm \frac{\pi}{3}, m \in Z$
Hence, $x=\frac{n \pi}{4} or, x=m \pi \pm \frac{\pi}{3}$ where m, n $\in$Z.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "(Each question 3 marks)" from TRIGONOMETRIC FUNCTIONS→TRIGONOMETRIC FUNCTIONS — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Solve the following linear inequations in R: $\frac{3\text{x}-2}{5}\leq\frac{4\text{x}-3}{2}$

If (n + 2)! = 60 [(n - 1)!], find n.

Let A = {x : x $\in$ N}, B = {x : x = 2n, n $\in$ N}, C = {x : x = 2n - 1, n $\in$ N} and D = {x : x is a prime natural number}. Find: $\text{B}\cap\text{D}$

Let A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}. Verify the following identities: $\text{A}\cap(\text{B}\cup\text{C})=(\text{A}\cap\text{B})\cup(\text{A}\cap\text{C})$

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{x}+2)^{\frac{5}{2}}-(\text{a}+2)^{\frac{5}{2}}}{\text{x}-\text{a}}$

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin(2+\text{x})-\sin(2-\text{x})}{\text{x}}$

Find the equation of the straight line which passes through the point P (2, 6) and cuts the coordinate axes at the point A and B respectively so that $\frac{\text{AB}}{\text{BP}}=\frac{2}{3}.$

Evaluate $\lim\limits_{\text{x}\rightarrow0}\text{f(x)},$ where $\text{f(x)}=\begin{cases}\frac{|\text{x}|}{\text{x}}, & \text{x} \ne0\\0, &\text{x} = 0\end{cases}.$

Prove that the product of the lengths of the perpendiculars drawn from the points $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ and $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to the line $\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$ is $b^2$.

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.