Advertisements
Advertisements
Questions
Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = π/2, x ≠ 0`
Solve the equation x: `sin^(-1) (5/x) + sin^(-1) (12/x) = π/2 (x ≠ 0)`
Advertisements
Solution 1
`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`
`sin^(-1) + cos^(-1) sqrt(1 - 144/x^2) = π/2`
Let `sin^(-1) 12/x = β`
`12/x = sin β = "OPP"/"HYP"`
`sqrt(x^2 - 144)/x = cos β = "Adj"/"HYP"`
`β = cos^(-1) (sqrt(x^2 - 144)/x^2)`
∴ `5/x = sqrt(1 - 144/x^2)`
`25/x^2 = 1 - 144/x^2`
`169/(x^2) = 1`
x2 = 169
x = 13
Solution 2
`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`
⇒ `sin^-1 (5/x) = π/2 - sin^-1 (12/x)`
⇒ `sin^-1 (5/x) = cos^-1 (12/x)` ...`[∵ sin^-1x + cos^-1x = π/2]`
⇒ `sin^-1 (5/x) = cos^-1 (12/x) = θ`
⇒ `sin θ = 5/x, cos θ = 12/x`
Since, sin2θ + cos2θ = 1
⇒ `(5/x)^2 + (12/x)^2 = 1`
⇒ `25/(x^2) + 144/(x^2) = 1`
⇒ `169/(x^2) = 1`
⇒ x = ±13
Since, x = –13 does not satisfy the given equation.
So, x = 13.
Notes
Students should refer to the answer according to their questions.
RELATED QUESTIONS
Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
y (1 + ex) dy = (y + 1) ex dx
Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.
Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.
x2 dy + y (x + y) dx = 0
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
(y2 − 2xy) dx = (x2 − 2xy) dy
Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
`dy/dx = log x`
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve: `("d"y)/("d"x) + 2/xy` = x2
The integrating factor of the differential equation `"dy"/"dx" (x log x) + y` = 2logx is ______.
lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
