solving trigonometric equations

In mathematics, an equation that contains trigonometric functions is known as trigonometric equations. For example: Cos u = ½. The roots of trigonometric function are obtained by the inverse trigonometric functions. Now see how to solving trigonometric equations. Suppose we have sin u + sin 2u + sin 3u = 0. This trigonometric equation can be reduced to form 2 sin 2u cos u + sin 2u = 0. In other word we can also write as: Sin 2u (2 cos u + 1) = 0. When we solve this equation we get the value of sin 2u as 0.  Sin 2u (2 cos u + 1) = 0;
⇨ Sin 2u = 0. And the value of cos u is:
⇨ 2 cos u + 1 = 0;
⇨ 2 cos u = -1;
⇨ Cos u = -1/2. So the value of cos u is -1/2. This equation gives the result of the trigonometric equations. So u = ½ arcsin 0 = n ⊼/2;
u = arcos (-1/2),
u = 2/3 ⊼ (3n + ½), Where, the value of ‘n’ is either positive or negative integer. Let’s discuss how to solve the trigonometric equations. Let we have the trigonometric equation 4 tan³ u – tan u = 0, that lies in the interval [0, 2⊼]. Then it can be solved as shown below:
So the given trigonometric equation is: 4 tan³ u – tan u = 0. We can write the equation as:
⇨ 4 tan³ u – tan u = 0;
⇨ Tan u (4 tan² u – 1) = 0, so the value of tan ‘u’ is 0;
Or tan u = + 1/√3. For every value of u ∈ [0, 2⊼],
⇨ Tan u = 0; It means the value of ‘u’ is 0, ⊼ or 2⊼.
While
Tan u = 1/√3,
u = ⊼ / 6 or 7⊼/ 6,
Tan u = -1/√3,
u = 5⊼ / 6 or 11⊼/ 6. This is how we can solve the trigonometric equation. Now we will see the Equation for Velocity. Equation of velocity is given by: Equation of velocity (v) = displacement (D) / time (T). The students of class 1st follow the icse syllabus for class 1.