Verified answer

... 2^(sin(x)+1) = √2

or 2^(sin(x)+1) = 2^(1/2)

or sin(x)+1 = 1/2

or sin(x) = -1/2

or x = Arcsin(-1/2)

or x = 210°, 330°

1a) 2sinx-cos^2x=2 Use the trig regulation that sin^2x + cos^2x = a million so: 2sinx - (a million-sin^2x) = 2 enhance the bracket: 2sinx - a million + sin^2x = 2 Make it right into a quadratic with the help of putting both on the different area: sin^2x + 2sinx - 3 = 0 To make it a lot less complicated, say: enable sinx = y and rewrite the equation: y^2 + 2y - 3 = 0 Factorise and be sure: (y - a million) ( y + 3) = 0 y - a million = 0 y + 3 = 0 y = a million y = -3 change sinx back into y: sinx = a million sinx = -3 x = ninety (calculator) x = no answer b) 2sin^2x + 5sinx - 3 = 0 Do an same because the top of the awesome one, enable sinx = y 2y^2 + 5y - 3 = 0 Factorise and be sure: (2y - a million) (y + 3) = 0 2y - a million = 0 y + 3 = 0 2y = a million y = -3 y = a million/2 change sinx back into y: sinx = a million/2 sinx = -3 x = 30 (calculator) x = no answer 2a) log<4>a million/4 placed it into what logs propose: 4^x = a million/4 x = -a million as 4^-a million = a million/4 b) log<8>512 placed it into what logs propose: 8^x = 512 x = 3 as 8^3 = 512 3a) 2log(x-a million) - log(3x-6) Use the indice log regulation so as that: log(x-a million)^2 - log(3x-6) Use the subtraction/branch log regulation so as that: log[ ((x-a million)^2) / (3x-6) ] b) 2log(2x+5) + 5log(x-8) Use the indice log regulation so as that: log(2x+5)^2 + log(x-8)^5 Use the addition/multiplication log regulation so as that: log[ (2x+5)^2 (x-8)^5 ] 4a) 9^(3x+a million) = 2^x placed it into log type: log9^(3x+a million) = log2^x Use the indice log regulation so as that: (3x+a million)log9 = xlog2 placed the x onto the different area, and the log9 onto the different area: (3x+a million) / x = (log2) / (log9) (3x+a million) / x = 0.315 (3sf) placed x onto the different area: 3x + a million = 0.315x placed the 0.315x onto the different area, and the a million onto the different area: 2.68x (3sf) = -a million x = -0.373 (3sf) b) 6(3^(4f-2)) = 2 placed it into log type: log6 + log3^(4f-2) = log2 Use the indice log regulation so as that: log6 + (4f-2)log3 = log2 placed the log6 onto the different area: (4f-2)log3 = log2 - log6 placed the log3 onto the different area with the help of dividing: 4f - 2 = (log2 - log6) / log3 4f - 2 = -a million 4f = a million f = a million/4

AS GIVEN , question MUST be read as :-

( 2^sin x) + 1 = √2

2^sin x = √2 - 1

sin x log 2 = log ( √2 - 1 )

sin x = log ( √2 - 1 ) / log 2

sin x = - 1.272

This cannot be so and would suggest that question should ACTUALLY be read as :-

2^(sin x + 1 ) = √2

( sin x + 1 ) log 2 = log √2

sin x (log 2) = (1/2) log 2 - log 2

sin x = (-1/2) log 2 / log 2

sin x = - 1/2

x = 7π/6 , 11π/6

TIP

Remember to use brackets.

Answerers should not have to guess how a question is meant to read.