Advertisements
Advertisements
Question
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
cos 2x = 1 − 3 sin x
Solution
1 – 2 sin2x = 1 – 3 sinx
2 sin2 x – 3 sin x = 0
sin x(2 sin x – 3) = 0
= sin x = 0 or 2 sin x – 3 = 0
sin x = 0 or sin x = `3/2`
sin x = `3/2` is not possible since sin x ≤ 1
∴ sin x = 0 = sin 0
The general solution is x = nit,
When n = 0, x = 0 ∉ (0°, 360°)
When n = 1, x = π ∈ (0°, 360°)
When n = 2, x = 2π ∉ (0°, 360°)
∴ The required solutions is x = π
APPEARS IN
RELATED QUESTIONS
If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]
If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]
Prove that:
\[\frac{\cos (2\pi + x) cosec (2\pi + x) \tan (\pi/2 + x)}{\sec(\pi/2 + x)\cos x \cot(\pi + x)} = 1\]
Find the general solution of the following equation:
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\sin x + \cos x = \sqrt{2}\]
Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]
Solve the following equation:
3sin2x – 5 sin x cos x + 8 cos2 x = 2
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.
If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]
Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`
Choose the correct alternative:
If cos pθ + cos qθ = 0 and if p ≠ q, then θ is equal to (n is any integer)
Choose the correct alternative:
If tan α and tan β are the roots of x2 + ax + b = 0 then `(sin(alpha + beta))/(sin alpha sin beta)` is equal to
