Question

if 3^2sin2a , 14 and 3^4-2sin2a are first thee terms of an AP for some a. Then 6th term of A.P. is

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Since

${3}^{2 \sin\theta - 1} ,14 \:, {3}^{4 - 2 \sin\theta }$

are in Arithmetic Progression

So

${3}^{2 \sin\theta - 1} + {3}^{4 - 2 \sin\theta } = 2 \times 14$

Let

$x = {3}^{2 \sin\theta }$

So above becomes

$\displaystyle \: \frac{x}{3} + \frac{ {3}^{4} }{x} = 28$

$\implies \: {x}^{2} - 84x + 243 = 0$

$\implies \: {x}^{2} - 81x - 3x+ 243 = 0$

$\implies \: x(x - 81) - 3(x - 81) = 0$

$\implies \: (x - 81)(x - 3) = 0$

So

$x = 81 \:, 3$

Now x = 81 gives

${3}^{2 \sin\theta } = 81$

$\implies \: {3}^{2 \sin\theta } = {3}^{4}$

$\implies \: {2 \sin\theta } = {4}$

$\implies \: \sin\theta = 2$

Which is impossible

Again x = 3 gives

$\: {3}^{2 \sin\theta } = {3}$

$\implies \: {2 \sin\theta } = 1$

Hence

$First \: Term = {3}^{2 \sin\theta - 1 } = {3}^{1 - 1 } = {3}^{0} = 1$

Second Term = 14

$Third \: Term = {3}^{4 - 2 \sin\theta } = {3}^{4 - 1} = {3}^{3} = 27$

So common difference = 14 - 1 = 13

So Sixth term

$= 1 + (6 - 1) \times 13$

$= 1 + 65$

$= 66$

Answer 2

$\blue{\huge{\underline{\underline{Answer:-}}}}$

Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.

We use the following formulae for solving the trigonometric equations:

·         sin q  = 0 Þ  q = np,

·         cos q = 0 Þq = (2n + 1),

·         tan q = 0  Þ  q =  np,

·         sin q = sin a  Þq =  np + (–1)na,         where aÎ [–p/2,  p/2]

·         cos q  = cos aÞq = 2np  ±  a, where aÎ [ 0, p]

·         tan q = tan a  Þ  q = np + a,                 where aÎ ( –p/2, p/2)

·         sin2 q = sin2 a , cos2 q = cos2 a,  tan2q = tan2 aÞq = np±a,

·         sin q = 1 Þq = (4n + 1),

·         cos q = 1 Þ  q = 2np ,

·         cos q = –1 Þ  q = (2n + 1)p,

·         sin q =  sin a  and  cos q =  cos aÞ  q = 2np + a.