For x -0- - then equation sin x-sin 2 x-sin 3 x-3 hasA- No solutionB- One solutionC- Infinitely many solutionsD- Three solutions

The correct option is A No solution sin ⁡x+2 sin ⁡2x sin⁡ 3x=3 ⇒ sin ⁡x+4 sin⁡x cos⁡x 3 sin ⁡x+4 sin^3⁡x=3 ⇒ sin ⁡x ( 2+4 cos ⁡x+4 sin^2⁡x )=3 ⇒ sin ⁡x ( 2+4 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Solving Simultaneous Trigonometric Equations

For x ∈0, π, ...

Question

For $x \in (0, π)$ , then equation $s i n x + s i n 2 x - s i n 3 x = 3$ has

Infinitely many solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Three solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

One solution

No worries! We‘ve got your back. Try BYJU‘S free classes today!

No solution

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
No solution

$s i n x + 2 s i n 2 x - s i n 3 x = 3 \Rightarrow s i n x + 4 s i n x c o s x - 3 s i n x + 4 s i n^{3} x = 3 \Rightarrow s i n x (- 2 + 4 c o s x + 4 s i n^{2} x) = 3 \Rightarrow s i n x (- 2 + 4 c o s x + 4 - 4 c o s^{2} x) = 3 \Rightarrow 2 + 4 c o s x - 4 c o s^{2} x = \frac{3}{s i n x} \Rightarrow 2 - (4 c o s^{2} x - 2 .2 c o s x + 1) + 1 = \frac{3}{s i n x} \Rightarrow 3 - {(2 c o s x - 1)}^{2} = \frac{3}{s i n x}$
LHS $\leq 3$ and RHS $\geq 3$
$∴$ Only possible scenario for a solution to exist is when LHS = RHS = 3. However, LHS and RHS cannot be simlultaneously equal to 3 for a particular value of $x$ . Hence no solution.

Suggest Corrections

For x∈(0,π), then equation sin x+sin 2x−sin 3x=3 has

For $x \in (0, π)$ , then equation $s i n x + s i n 2 x - s i n 3 x = 3$ has