Question

If sin 3x = 1 and 0° < 3x < 90°, find the values of
(i) sin x
(ii) cos 2x
(iii) tan^2 x - sec^2 x.​

Answer 1

Step-by-step explanation:

sin 3x = 1

3x = 90

x = 30

Now

sin 30 = 1/2

cos 2×30 = cos 60 = 1/2

tan^2 x - sec^2 x = -1

Hope it helps you mate.

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Answer 2

Given,

• sin 3x = 1
• 0° < 3x < 90°

To find,

• sin x
• cos 2x
• $tan^2x -sec^2x$

Solution,

If sin 3x = 1 and 0° < 3x < 90°, then the values of sinx is 1/2, cos2x is 1/2, and $tan^2x -sec^2x$is -1.

We can simply find the values of the given trigonometric functions by using the concepts of trigonometry,

                Sin 3x = 1

As we know, sin 90° = 1, then replacing 1 by sin 90° in the above equation, we get

                Sin 3x = Sin 90°

Pre-multiplying by Sin⁻¹ on both sides, we get

       Sin⁻¹ Sin 3x = Sin⁻¹ Sin 90°

                      3x = 90

                       x = 30

 Now, sinx = sin 30° = 1/2

    For cos 2x,

              Cos 2x = cos 2(30)

                           =   cos 60°   [ cos 60° = 1/2]

                           = 1/2

              Cos 2x = 1/2

Now, $tan^2x -sec^2x$

       =   tan² 30 - sec² 30

       =    1/3   - 4/3          [tan 30° = 1/√3]  [ sec 30° = 2/√3]

       =        1-4/3

       =          -3/3

       =           -1

$tan^2x -sec^2x$ = -1

Hence, if sin 3x = 1 and 0° < 3x < 90°, then the values of sinx is 1/2, cos2x is 1/2, and $tan^2x -sec^2x$ is -1.